Number Theory James Emery Last Addition or Edit: 6/7/2014
Contents 1 Introduction
3
2 Prime Numbers
3
3 The Prime Number Theorem
4
4 The Euclidean Algorithm for the Greatest Common Divisor
5
5 Python gcd.py Greatest Common Divisor Program
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6 The Greatest Common Divisor Example in Java
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7 The Greatest Common Divisor Example in Scheme
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8 The Greatest Common Divisor Example in Python 17 8.1 Example: Greatest Common Divisor . . . . . . . . . . . . . . 18 9 Modular Arithmetic
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10 The Euler Totient Function
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11 Decimal Expansions
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12 Fermat’s Theorem
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13 Fermat’s Little Theorem
28 1
14 A Fast Exponential Algorithm for Integers
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15 A Scheme Program to Test For a Number Being Prime
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16 A Python Program to Test For a Number Being Prime
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17 Carmichael Numbers
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18 Fermat’s Last Theorem
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19 Mersenne Primes
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20 The Chinese Remainder Theorem
40
21 Wilson’s Theorem
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22 Hardy’s Cab Number
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23 Quadratic Residues
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24 Diophantine Equations
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25 Goldbach’s Conjecture
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26 The Riemann Zeta Function
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27 The Riemann Hypothesis
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28 The Sieve of Eratosthenes
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29 Perfect Numbers
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30 Twin Primes
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31 Analytic Number Theory
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32 Repeating Decimals
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33 Appendix, Primes Less Than 10000
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2
34 Fibonacci Numbers
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35 Carl Fredich Gauss. The Prince of Mathematicians
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36 Bibliography
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1
Introduction
The Theory of Numbers is that part of mathematics that studies properties of the counting numbers and the integers. It has generated many of the concepts of modern Abstract Algebra.
2
Prime Numbers
A prime number is a positive integer that cannot be written as the product of two integers, each greater than 1. The 25 primes less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Any two positive integers n and m, have a greatest common divisor. This is written as (m, n). If (m, n) = 1, the numbers are called relatively prime. This means they have no common factor that is greater than 1. Notice that by the definition every number is relatively prime to 1. The Fundamental Theorem of Arithmetic says that every positive integer greater than 1, can be written as a product of powers of primes in a unique way. Thus for example 45 = (2)(32 ), and 59424750 = (2)(32 )(53 )(74 )(11). The number 1 is not considered a prime, even though it cannot be written as a product of two integers each greater than 1. There are an infinite number of primes. For suppose there are a finite number of them. Multiply these finite numbers together and add 1. This 3
number is not divisible by any of the finite primes. But a number is either divisible by a prime, or a prime itself. This is a contradiction. Therefore there are an infinite number of primes. This argument was given by Euclid.
3
The Prime Number Theorem
The prime number theorem says that the number of primes less than natural number n is approximately equal to n . ln(n) The Prime Number Theorem. If π(n) is the number of primes less than n, then the ratio of π(n) to n/ ln(n) goes to 1 as n goes to ∞, lim
n→∞
ln(n) π(n) n
!
= 1.
The following table suggests that this is true. n 101 102 103 104 105 106 107 108 109 1010
π(n) n/ ln(n) ratio 4 4.3 0.93 25 21.7 1.15 168 144.8 1.16 1,229 1,086 1.13 9,592 8,686 1.10 78,498 72,382 1.08 664,579 620,420 1.07 5,761,455 5,428,681 1.06 50,847,534 48,254,942 1.05 455,052,511 434,394,482 1.048
A proof can be found in the book on Analytic Number Theory by Tom Apostle.
4
4
The Euclidean Algorithm for the Greatest Common Divisor
Given two positive integers b1 and b2 , we shall show how to compute the gcd. Let us assume that b1 > b2 . We can divide a larger integer by a smaller one getting a quotient and a remainder. So suppose b1 = 108 and b2 = 30. Dividing 108 by 30 we get a quotient of 3 with a remainder of 18. The remainder 18 is less than the divisor 30. We can write this as 108 = 3 · 30 + 18, and . The remainder becomes the next divisor in a continuing sequence: 30 = 1 · 18 + 12. 18 = 1 · 12 + 6 12 = 2 · 6 + 0. When we get a zero remainder, the previous remainder is the gcd. So gcd = 6. Next we shall prove this algorithm always produces the gcd. So we start with b1 = k1 b2 + b3 , where k1 and b3 are integers. Number k1 is the quotient, b2 is the divisor, and b3 is the remainder. A divisor is always greater than a remainder, so b2 > b3 . We may continue this division b1 = k1 b2 + b3 , b2 = k2 b3 + b4 b3 = k3 b4 + b5 .................... bn−1 = kn−1 bn + bn+1 bn = kn bn+1 + bn+2 , until the remainder bn+2 is zero, when the division stops. We have a strictly decreasing sequence of integers b1 > b2 > b3 > .... > bn+2 . 5
Eventually we must reach a step n where the remainder bn+2 is zero, where we can’t continue the division. At this point we claim that g = bn+1 is the greatest common divisor of the integers b1 and b2 . Going backwards, we have bn = kn bn+1 , bn−1 = kn−1 bn + bn+1 , bn−2 = kn−2 bn−1 + bn , .................... b1 = k1 b2 + b3 , showing that bn+1 eventually divides b2 and b1 . Explicitly from the first equation bn = kn bn+1 , we see that bn+1 divides bn . Continuing down the list of equations, we see that for each k, bk+1 divides bk , and thus ultimately bn+1 divides b2 and b1 . So g = bn+1 is a common divisor of b2 and b1 . Next we must prove that it is the largest common divisor. So suppose h is any other common divisor. That is suppose that h divides b1 and b2 . Then from the first equation of the first list above, namely b1 = k1 b2 + b3 , h must divide b3 , and so from the second equation it must divide b4 , and so on, down the list, and so h must ultimately divide bn+1 . Hence bn+1 ≥ h. Therefore g = bn+1 is the gcd of m and n. Note. Our assumption that m = b1 > b2 = n in the calculation is not important, because if m=b1 < b2 = n the first step will be b1 = k1 b2 + b3 = 0 · b2 + b3 , that is b3 = b1 , so the second step in effect is restarting the calculation, reversing the positions of b1 and b2 . 6
The following proposition is a very important one in number theory and is often used in proofs. Proposition. Given two integers, b1 ≥ 2 and b2 ≥ 2, there exists two integers α1 and β1 such that the greatest common divisor g is given by g = α1 b1 + β1 b2 . Proof. The Euclidean algorithm presented above for the gcd(b1 ,b2 ) = g is b1 = k1 b2 + b3 , b2 = k2 b3 + b4 b3 = k3 b4 + b5 .................... bn−2 = kn−2 bn−1 + bn bn−1 = kn−1 bn + g bn = kn bn+1 . Solving the next to last equation, equation n − 1, for g we have g = bn−1 − kn−1 bn We can write this as g = αn−1 bn−1 + βn−1 bn , where αn−1 = 1, and βn−1 = −kn−1 . Now consider the division step n − 2, where equation n − 2 is bn−2 = kn−2 bn−1 + bn . Solving for bn , we have bn = bn−2 − kn−2 bn−1 7
Substituting for bn in g = αn−1 bn−1 + βn−1 bn , we have g = αn−1 bn−1 + βn−1 (bn−2 − kn−2 bn−1 ) = βn−1 bn−2 + (αn−1 − βn−1 kn−2 )bn−1 = αn−2 bn−2 + βn−2 bn−1 , where αn−2 = βn−1 and βn−2 = αn−1 − βn−1 kn−2 This relationship will be repeated, so that at the pth step we find g = αp bp + βp bp+1 , where αp = βp+1 and βp = αp+1 − βp+1 kp . So finally g = α1 b1 + β1 b2 . Thus by saving the quotients k1 , k2 , ...kn−1 we can construct the numbers α1 and β1 . Corollary. If integers m and n are relatively prime, then m mod(n) has a multiplicative inverse mod(n). Proof. We have integers α and β so that 1 = αm + βn, So that α is the multiplicative inverse of m in the ring Zn , that is αm = 1 mod(n), 8
because a multiple of n is zero in the modular ring Zn . Corollary If p is a prime then the modular ring Zp is a field (Every nonzero element has a multiplicative inverse). There is a java program gcdex.java in the next section for the calculation of the gcd, and the numbers α and β. If the gcd =1, it calculates the multiplicative inverse of m mod n. We can find α and β by hand. At the beginning of this section we calculated 6 as the gcd of 108 and 30 as follows: 108 = 3 · 30 + 18, 30 = 1 · 18 + 12, 18 = 1 · 12 + 6, 12 = 2 · 6 + 0. So from the third line the gcd is 6 = 18 − 12 We use the second equation to replace 12 by a combination of 18 and 30, 12 = 30 − 18 giving the gcd as 6 = 18 − (30 − 18) = 2 · 18 − 30. Then we use the first line to replace 18 as a combination of 108 and 30 18 = 108 − 3(30), giving the gcd as 6 = 2(108 − 3 · 30) − 30 = 2 · 108 − 7 · 30. So α = 2, and β = −7. Corollary If an integer k divides the integers m and n, then k divides their greatest common divisor (m, n). Proof. This follows because there exists integers α and β such that 9
(m, n) = αm + βn. Proposition. If r is the remainder on division of a by b, then (a, b) = (b, r). Proof. Suppose a = kb + r. From this (b, r) divides a. Because it also divides b, it thus divides (a, b). We have r = a − kb, and so because (a, b) divides both a and b, it divides r. Therefore (a, b) divides both b and r, so (a, b) divides (b, r). Obviously two positive integers that divide each other are equal. It follows that (a, b) = (b, r)
5
Python gcd.py Greatest Common Divisor Program
This program computes the gcd(m,n) of two integers m and n, and computes numbers α and β such that gcd(m, n) = αm + βn Running the program: Example 1. c:\je\py>python gcd.py 510510 27170 m= 510510 n= 27170 510510 = 18 * 27170 + 21450 27170 = 1 * 21450 + 5720 21450 = 3 * 5720 + 4290 5720 = 1 * 4290 + 1430 4290 = 3 * 1430 + 0 10
v = [gcd,alpha,beta]= [1430, -5, 94] gcd(m,n)= 1430 alpha*m+beta*n= 1430 So the two numbers have the factor 1430 in common. 1430 = (11)(13)(5)(2), Example 2. Here is another example showing that the integers 946520984 and 87905179 are relatively prime. c:\je\py>python gcd.py 946520984 87905179 m= 946520984 n= 87905179 946520984 = 10 * 87905179 + 67469194 87905179 = 1 * 67469194 + 20435985 67469194 = 3 * 20435985 + 6161239 20435985 = 3 * 6161239 + 1952268 6161239 = 3 * 1952268 + 304435 1952268 = 6 * 304435 + 125658 304435 = 2 * 125658 + 53119 125658 = 2 * 53119 + 19420 53119 = 2 * 19420 + 14279 19420 = 1 * 14279 + 5141 14279 = 2 * 5141 + 3997 5141 = 1 * 3997 + 1144 3997 = 3 * 1144 + 565 1144 = 2 * 565 + 14 565 = 40 * 14 + 5 14 = 2 * 5 + 4 5 = 1 * 4 + 1 4 = 4 * 1 + 0 v = [gcd,alpha,beta]= [1, 18825831, -202707557] gcd(m,n)= 1 alpha*m+beta*n= 1
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So these numbers m and n are relatively prime, and α = 18825831 is the multiplicative inverse of m in the ring Zn , because any multiple of n is zero in the ring Zn . Listing: #gcd.py greatest common divisor, getting m and n from the command line # May 30, 2014 #see gcdexample.pi , where m and n are defined inside the program. import sys #function greatest common divisor def gcd(b1,b2): b=[] k=[] rm=[] v=[] v=[1,2,3] b.append(0) b.append(b1) b.append(b2) k.append(0) rm.append(0) n=0 while (b2 > 0): n=n+1 #print " b1= ",b1 #print " b2= ",b2 g=b2 q=b1/b2 k.append(q) #note, if b2>b1 then q=0, then r=b1 so b1 and b2 are switched #print " q= ",q r=b1-b2*q; b.append(r) print " ",b1,"=",q,"*",b2,"+",r rm.append(r) #print " r= ",r b1=b2 12
b2=r v[0]=g #print #print " number of divisions n=",n #print " b=" #for i in range(1,len(b)): # print " b[",i,"]=",b[i] #print " length of k=",len(k) #print " quotients=",k #for i in range(1,len(k)): # print " k[",i,"]=",k[i] #print " remainders=",rm #for i in range(1,len(rm)): # print " rm[",i,"]=",rm[i] alpha=1 beta=-k[n-1] #print " g=alpha*b[n-1]+beta*b[n]=",alpha*b[n-1]+beta*b[n] p=n-2 while (p>0): tmp=alpha alpha=beta beta=tmp-beta*k[p] #print "p=",p #print " g=alpha*b[p]+beta*b[p+1]=",alpha*b[p]+beta*b[p+1] p=p-1 #print #print " alpha=",alpha," beta=",beta #print " g=alpha*b[1]+beta*b[2]=",alpha*b[1]+beta*b[2] v[1]=alpha v[2]=beta return v #function least common multiple def lcm(m,n): v=m*n/gcd(m,n) return v
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#main #print sys.argv #print sys.argv[0] na=len(sys.argv) #print ’number of arguments = ’, na if( na < 3): print "gcd.py, greatest common divisor of two integers" print "Usage: python gcd.py m n " sys.exit(0) m=int(sys.argv[1]) print " m= ",m n=int(sys.argv[2]) print " n= ",n v=gcd(m,n) #print " length v = ",len(v) print " v = [gcd,alpha,beta]=",v print " gcd(m,n)= ",v[0] alpha=v[1] #print " alpha=",alpha beta=v[2] #print " beta=",beta print " alpha*m+beta*n= ",alpha*m+beta*n
6
The Greatest Common Divisor Example in Java
Here is a java program for the calculation of the gcd, the numbers α and β If the gcd =1, it calculates the multiplicative inverse of m mod n. //gcdex.java, greatest common divisor of two integers 1-1-2008 class gcdex{ public static void main(String args[]) { int b[]; int a[]; int na;
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int m; int n; int g; a = new int[2]; b = new int[10];
na=args.length; if(args.length < 1){ System.out.println(" greatest common divisor of two integers " ); System.out.println(" Usage: java gcd m n " ); return; } for(int i=0;i
b1; b2; b3; na; nk; ns; r;
int alpha; int beta; int g; int j; int tmp; nk=100;
15
k = new int[nk]; b1=m; b2=n; tmp=b1/b2; r=b1-tmp*b2; if(r == 0){ System.out.print(" gcd= " alpha=0; beta=1; g=alpha*b1+beta*b2;
+ b2 + "\n");
return g; } System.out.print(" b1= " + b1 + " b2= " + b2 + "\n"); ns=0; while(r>0){ k[ns]=b1/b2; r=b1-k[ns]*b2; System.out.print(" k[" + ns + "] = " + k[ns] + " r= " + r + "\n"); if(r > 0){ b1=b2; b2=r; System.out.print(" b1= " + b1 + " b2= " + b2 + "\n"); ns=ns+1; } } g=b2; j=ns-1; alpha=1; beta=-k[j]; j=j-1; while(j >= 0){ tmp=alpha; alpha=beta; beta=tmp - k[j]*beta; j=j-1; } a[0]=alpha; a[1]=beta; return g; } }
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7
The Greatest Common Divisor Example in Scheme
Application. To compute the greatest common divisor (a, b),of two integers a and b, we can compute the remainder r of a/b and compute (b, r), recursively. Stopping when the value of b reaches zero. Here is a Scheme function to do this calculation: (define (gcd1 a b) (if (= b 0) a (gcd1 b (remainder a b)))) We have used gcd1 as the name of the function, because there is an actual gcd function built into Scheme.
8
The Greatest Common Divisor Example in Python
This program computes gcdexample(m,n) where the two integers are defined in the program. See also program gcd.py where the integers are defined as command line parameters. The python program computes with long integers. #gcdexample.py greatest common divisor, m and n defined in program main #also see command line version gcd.py, getting m and n from command line parameters import sys # #function greatest common divisor def gcd(m,n): b1=m b2=n while (b2 > 0): v=b2 q=b1/b2 #note, if b2>b1 then q=0, then r=b1 so b1 and b2 are switched r=b1-b2*q; b1=b2 b2=r return v #function least common multiple def lcm(m,n): v=m*n/gcd(m,n) return v
17
#program main m=2*3*5*7*11*13*14*15*200*189*3574 n=91*93*101*3*7*35789*1000*13*14 print " m=",m print " n=",n k=gcd(m,n) print " gcd(m,n)= ",k j=lcm(m,n) print " lcm(m,n)= ",j mdgcd=m/k ndgcd=n/k print " m=m/gcd =",mdgcd print " n=n/gcd =",ndgcd d=gcd(mdgcd,ndgcd) print " gcd(m,n) = ",d
8.1
Example: Greatest Common Divisor
#gcd.py greatest common divisor import sys # #function greatest common divisor def gcd(m,n): b1=m b2=n while (b2 > 0): # print " b1= ",b1 # print " b2= ",b2 v=b2 q=b1/b2 # print " q= ",q r=b1-b2*q; # print " r= ",r b1=b2 b2=r return v #function least common multiple def lcm(m,n): v=m*n/gcd(m,n) return v #print sys.argv #print sys.argv[0] na=len(sys.argv) #print ’number of arguments = ’, na if( na < 3): print "gcd.py, greatest common divisor of two long integers" print "Usage: python gcd.py m n " sys.exit(0) m=int(sys.argv[1])
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#print " m= ",m n=int(sys.argv[2]) #print " n= ",n k=gcd(m,n) print " gcd= ",k j=lcm(m,n) print " lcm= ",j mdgcd=m/k ndgcd=n/k print " m=m/gcd =",mdgcd print " n=n/gcd =",ndgcd d=gcd(mdgcd,ndgcd) print " gcd = ",d
Running the gcd.py program: $ python gcd.py 200560490130 3451796661527379 m= 200560490130 n= 3451796661527379 gcd(m,n)= 9582441 lcm(m,n)= 72246104125768042470 m= m/gcd = 20930 n= n/gcd = 360221019 gcd(m,n) = 1
9
Modular Arithmetic
Let m and n be integers. By definition m mod(n) is the remainder of m/n. So if m = qn + k, with k < n, we write m mod(n) = k. The integers can be separated into equivalence classes with the equivalence relation (a, b) (we write also a ∼ b), where a and b are equivalent iff there exist an integer k such that a − b = kn. 19
That is, a is equivalent to b iff a and b differ by a multiple of n. Then a is equivalent to b if a mod(n) = b mod(n). This is an equivalence relation defined to be a relation that satisfies the following three properties. (1) (a,a), since a − a = 0n (2) If (a,b), then a − b = kn, so b − a = −kn thus (b,a). (3) If (a, b) and (b, c) then there exists a k1 and a k2 so that a − b = k1 n and b − c = k2 n. Thus a − c = (a − b) + (b − c) = k1 n + k2 n = (k1 + k2 )n. So ( a,c) Every integer is equivalent to exactly one integer between 0 and n − 1. Thus for example ((a + b)0 + c)0 ∼ (a + b + c)0 ∼ (a + (b + c)0 )0 , where by (a + b)0 we mean addition mod n. So addition in In is associative. Similar arguments show that the integers mod n, In , form a ring. The integers mod n are commonly written as either Zn or In . Here is an addition table for the integers mod 5. + 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
Here is a multiplication table for the integers mod 5 × 0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1 20
Proposition An integer m in In , has a multiplicative inverse iff m and n are relatively prime. Proof As a consequence of the Euclidean algorithm, the GCD of n and m can be written as GCD(n, m) = αm + βn, for some integers α and β. So if n and m are relatively prime then there exists α and β so that 1 = αm + βn. But this means that α is the multiplicative inverse of m in In . On the other hand, if α is the multiplicative inverse of m, then there exists a β so that 1 = αm + βn. If m and n had a factor greater than 1 in common, say k, then we would have the fraction 1 m n = α + β = αm1 + βn1 , k k k equal to an integer. Thus m and n are relatively prime. Now every integer 1, 2, 3, ..., n − 1 is relatively prime to n if and only if n is prime. Thus the ring In is a field iff n is prime. Here is an addition table for the integers mod 10. + 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 6 7 8 9 0
2 2 3 4 5 6 7 8 9 0 1
3 3 4 5 6 7 8 9 0 1 2
4 4 5 6 7 8 9 0 1 2 3
5 5 6 7 8 9 0 1 2 3 4
6 6 7 8 9 0 1 2 3 4 5
7 7 8 9 0 1 2 3 4 5 6
8 8 9 0 1 2 3 4 5 6 7
9 9 0 1 2 3 4 5 6 7 8
Notice that the following multiplication table for the integers mod 10, shows that the numbers 2, 4, 5, 6, 8, which are not relatively prime to 10, do not have multiplicative inverses. For example, in the row for multiplication by 2, 21
we see there is no occurrence of 1. This means nothing times 2 gives 1, and so 2 has no inverse. Here is a multiplication table for the integers mod 10. × 0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9
2 0 2 4 6 8 0 2 4 6 8
3 0 3 6 9 2 5 8 1 4 7
4 0 4 8 2 6 0 4 8 2 6
5 0 5 0 5 0 5 0 5 0 5
6 0 6 2 8 4 0 6 2 8 4
7 0 7 4 1 8 5 2 9 6 3
8 0 8 6 4 2 0 8 6 4 2
9 0 9 8 7 6 5 4 3 2 1
The numbers that are relatively prime to 10, namely 1, 3, 7, 9 do form a multiplicative group, because they do all have multiplicative inverses. And also if two numbers are each relatively prime to third number m, then the product is also relatively prime to m. This is because the product still has no factors in common with m. However, this subset of numbers 1, 3, 7, 9 does not form a group under addition. For example 1 + 1 = 2, which not in the subset 1, 3, 7, 9. The φ function is the number of relative primes to 10, that are less than 10, φ(10) = 10(1 − 1/2)(1 − 1/5) = 4, 2 and 5 being the primes that divide 10. These numbers 1, 3, 7, 9 are called ∗ the reduced residue system mod 10, and are written Z10 , and form a multiplicative group, that is they have inverses and are closed. If m and n are equal mod(k), then m − n is divisible by k. If j is any integer, then jm and jn are equal mod(k), because the difference jm − jn = j(m − n) is a multiple of k. But the opposite is not true, if jm and jn are equal mod(k), then m and n are not necessarily equal mod(k). This can be seen from the multiplication table above. For example, 6 × 7 = 2 and 6 × 2 = 2, yet 7 does not equal 2. If k is a prime, then Ik is a field, and the cancellation law does hold. The cancellation law does not hold in general for Ik , the ring of integers mod 22
k. It does hold if k is prime. Proposition. If r is the remainder on division of a by b, then (a, b) = (b, r). Proof. Suppose a = kb + r. From this (b, r) divides a. Because it also divides b, it thus divides (a, b). We have r = a − kb, and so because (a, b) divides both a and b, it divides r. Therefore (a, b) divides both b and r, so (a, b) divides (b, r). Obviously two positive integers that divide each other are equal. It follows that (a, b) = (b, r)
10
The Euler Totient Function
The word totient comes from a Latin word meaning ”so many”, and is pronounced to rhyme with quotient. The Euler Totient function, also called the φ(n) function is a function on the set of positive integers, so that φ(n) is the number of positive integers less than n, that are relatively prime to n. Recall that two integers are relatively prime if their gcd, (m, n) = 1. If p is a prime, then (1, p) = 1, so 1 is counted by the φ function. If p is prime then no numbers less than p have a prime factor that is a prime factor of p. Therefore we have the following theorem. Theorem. If p is prime, then φ(p) = p − 1. Theorem. If p is a prime then φ(pk ) = pk−1 (p − 1). Proof. This follows because if we list all of the integers from 1 to p we have 1, 2, 3, ..., p, p + 1, p + 2, ..., 2p, ..., pk So one has repeatedly p − 1 integers that are equivalent to 1, 2, 3, 4, ..., (p − 1) mod(p), followed by a multiple of p. This repeating occurs pk /p = pk−1 times. 23
So all of these integers are relatively prime to pk except for the multiples of p. It follows that there are pk−1(p − 1) of these numbers. So φ(pk ) = pk−1 (p − 1). So for example consider p = 32 . The list of numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9. So the relatively prime numbers to 32 = 9, skipping the multiples of 3, are 1, 2, 4, 5, 7, 8 So φ(32 ) = 31 (3 − 1) = 6. We write j|m to mean integer j divides integer m. Theorem. If j|m, j|k and k = `mod(m), then j|`. Proof. If j|m and j|k, then for some q1 , q2 m = q1 j and k = q2 j. If k = `mod(m), then for some q3 ` = k + q3 m = q2 j + q3 q1 j. Corollary. If k is relatively prime to m and ` = kmod(m), then ` is relatively prime to m. Theorem. If m and n are relatively prime, then φ(mn) = φ(m)φ(n). Proof. Display the numbers 1, 2, 3, 4, ....., mn as an m by n matrix A, the first row being 1, 2, 3, 4.....n the second n + 1, n + 2, n + 3, n + 4.....2n 24
and so on. In the first row, φ(n) of these numbers are relatively prime to n. Consider the element A(1, k) in the first row that is relatively prime to n. Every element of the kth column is equal mod(n) to A(1, k) by the nature of matrix A. By the preceding corollary, every element of the kth column of A is relatively prime to n. In a given column k, the elements A(1, k), A(2, k), ..., A(m, k) are distinct mod(mn), because m and n are relatively prime allowing application of the cancellation law. So they are equivalent to 1, 2, 3, 4, 5, 6, ..., m mod(m). Therefore there are exactly φ(m) in each column that are relatively prime to m. So there are φ(m)φ(n) relatively prime to both m and n. Note. Obviously this result does not hold if m and n are not relatively prime. For example let m = 4 and n = 6 then φ(m) = 2 and φ(n) = 1, but φ(24) = 8. Corollary If n=
k Y
pj j ,
k Y
φ(pj j ).
n
j=1
where p1 , p2 , ...pk are primes, then φ(n) =
n
j=1
=
k Y
n −1
pj j
(pj − 1).
j=1
=n
Qk
j=1 (pj − Qk j=1 pj
φ(n) = n
1)
.
1 (1 − ). p p|n
Y
The last product is over all of the primes p that divide n. The equation Y 1 φ(n) = n (1 − ). p p|n is derived directly in Apostol using the Mobius function.
25
11
Decimal Expansions
Consider the number 77/13 = 5.923076923076923076923076923076 where the infinitely repeating digits are underlined. These repeating digits form the number p = 923076. We can write the number as an integer plus a number less than 1. 77/13 = 5 + 12/13 This number can be written as a continued fraction 77/13 = [5, 1, 12] = 5 + 1/(1 + 1/(12)) A continued fraction that is written as [a1 , a2 , a3 , a4 , ..., an ] means a1 + 1/(a2 + 1/(a3 + 1/(a4 + 1/(......)))) Let x = 12/13. Then x = .923076923076923076923076923076923076923076 1000000x = 923076 + x. So 999999x = 923076 So that x = 923076/999999. We have gcd(923076, 999999) = 76923.
26
Dividing by the gcd, we reclaim the fraction x = 12/13. Using this method of finding the rational number corresponding to an infinitely repeating decimal, we will always find a rational number with denominator whose digits are all nines. So it must be true that every number n divides some number of the form |999...9 {z }
= 10r − 1,
r
for some r. To show this let m = n − 1. So m/n is a repeating decimal less than 1 of say r digits, with the first digit > 1 . Let a number p be the integer formed by the repeating digits. Let m/n = x, then 10r x = p + x Then x = p/(10r − 1) = m/n So np = m(10r − 1) So n divides m(10r − 1), but n does not divide m, so n divides (10r − 1). Notice that for a rational number m/n, the number r of repeating digits in its decimal expansion must be less than n. This follows because in calculating each digit in its decimal expansion there will be a remainder that can be zero or a number between 1 and n−1. As soon as one of these remainders repeats then the digits will repeat. so there can be only n − 1 digits before a repeat. The maximum number of repeating digits can occur as in 1/7 = 0.142857142857142857142857142857, where there are 6 repeated digits.
12
Fermat’s Theorem
Fermat’s theorem says that if two integers a and m are relatively prime, a raised to the power of φ(m) is equal to 1 mod(m). 27
Fermat’s Theorem. If a and m are relatively prime, then aφ(m) = 1 mod(m). Proof Let R be the reduced residue multiplicative group consisting of the elements of Im that are relatively prime to m. There are φ(m) elements of R, say R = {r1 , ..., rφ(m)} . Because a and m are relatively prime, a ∈ R. Hence R = {ar1 , ..., arφ(m)} . This follows because every element of R has an inverse so that if ari = arj then ri = rj , so i = j. Hence for each i there exists a j so that ari = rj . Multiplying the φ(m) such equations together, we get aφ(m) r1 r2 ...rφ(m) = r1 r2 ...rφ(m) , from which we get that aφ(m) = 1 mod(m). Example. Let a = 9 and m = 100. Then φ(100) = 100(1 − 1/2)(1 − 1/5) = 100(1/2)(4/5) = 40. aφ(m) = 940 = 147808829414345923316083210206383297601, which is 1 mod 100.
13
Fermat’s Little Theorem
Fermat’s little theorem is a Corollary to Fermat’s theorem. If m is a prime in Fermat’s Theorem then we get Fermat’s Little Theorem. Fermat’s Little Theorem. If a and p are integers, and p is a prime, then ap = a mod(p). Proof Because p is a prime, φ(p) = p − 1 so ap−1 = 1 mod(p). Multiplying both sides by a, we get ap = a mod(p). 28
14
A Fast Exponential Algorithm for Integers
Consider computing an integer k raised to the n power, k n . The obvious algorithm is simply to multiply k times itself n − 1 times, as in 25 = 2 ∗ 2 ∗ 2 ∗ 2 ∗ 2 = 32. However, consider the case that k is 200 digits long and n is a thousand digits long. The obvious calculation will take quite some time. Let us write this exponential function as f (k, n) = k n . Consider the product operator ∗ for multiplication in what is called infix notation, then the product of b times c is b ∗ c. By infix we mean that the operator appears between the numbers, which is the normal way of writing arithmetic expressions. In prefix notation this product would be written as ∗bc, where the operator comes first followed by the two numbers, which are to be multiplied. Let us use the letter p as the multiplication operator, s as the square operator, and e as the exponential operator. Then for example in our prefix notation p98 = (9)(8) = 72, s6 = 62 = 36, , and 35 = e3 5 = 243. We can decompose the exponent 5 into two pieces so that 35 = 3(32 )(32 ) = 3 ∗ (32 ) ∗ (32 ). In our prefix notation this would be written p3ss3. We evaluate this step by step from the right, applying the operators in succession p3ss3 = p3s9 = p3 81 = 243. Now our fast exponential algorithm will consist in decomposing the exponential operator into a prefix stack consisting of just a sequence of symbols, a string made up of a combination of the individual symbols p, s, k, where k 29
is the number to be raised to the nth power, p is the multiplication operator, and s is the square operator. Now we convert e(k, n) → ekn, into an expression involving p, s, e, where the new e expression has a smaller exponent. In fact we divide the exponent by 2 at each step. There are two cases, namely where n is even, and where n is odd. In the case that n is even, we translate k n → ekn to k n/2 k n/2 = sek(n/2) So for example if n = 1024, then k 1024 = ek(1024) = sek(512) The parentheses are there to indicate that anything surrounded by parentheses is a single symbol, so (n/2) is the actual number obtained immediately, not three symbols n, , and 2, and (1024) means the single number 1024 not the four numbers 1,0,2,4. That is, the division operator does not appear in the symbol stack, this division by two is done immediately before the result is placed in the stack. In the case that n is odd, 2 does not divide n evenly so we have a remainder of 1. So we need to multiply additionally by the number k, so ekn = pksek(n/2). So for example if n = 1023 we have k 1023 = ek(1023) = pksek(511). Now we continue this conversion until we reach an exponential ek0 = k 0 = 1 Then we evaluate our expression in the stack from right to left obtaining the value of k n , with a very very much smaller amount of arithmetic than in the simple minded algorithm of multiplying k by itself n − 1 times. A Scheme version of this function, which is recursive, that is, it calls itself, is used in the next section in a prime test. It is 30
(define (fast-exp b n) (cond ((= n 0 ) 1 ) ((even? n) (square (fast-exp b (/ n 2)))) (else (* b (fast-exp b (- n 1)))))) The line with the ”cond” word in it is a kind ”if” command, which here returns 1 if the exponent is zero, or else goes to the next two lines handling division by 2 and the recursive call. Note that the conversion of the calculation, to a stack expression, and then its evaluation right to left, is handled by the scheme interpreter itself.
15
A Scheme Program to Test For a Number Being Prime
This Scheme program comes from the book, Abelson, Harold and Sussman, Gerald J., Structure and Interpretation of Computer Programs. See Fermat’s Theorem on p89 of Elementary Theory of Numbers, by Harriet Griffin, McGraw-Hill, 1954. If a and m are relatively prime then aφ(m) = 1mod(m), where φ is the Euler φ functions, the number of integers less than m and relatively prime to m. If p is prime then φ(p) = p − 1, so ap−1 = 1mod(p) or ap = amod(p). This is Fermat’s little theorem. See file fermat-test.cs in the scheme directory. Here are the Scheme functions for the Fermat test of a prime using Fermat’s little theorem, Abbelson and Sussman p42. So (fermat-test n) returns #t if n is a prime, #f if n is not a prime The Function fermat-test uses subfunctions: square (square x), squares x fast-exp, (fast-exp b n), raises b to the nth power expmod, (expmod b e m), raises b to power e mod m Definitions: (define (square x) ( * x x))
31
(define (fast-exp b n) (cond ((= n 0 ) 1 ) ((even? n) (square (fast-exp b (/ n 2)))) (else (* b (fast-exp b (- n 1)))))) (define (expmod b e m) (cond ((= e 0) 1 ) ((even? e) (remainder (square (expmod b (/ e 2) m)) m)) (else (remainder (* b (expmod b (- e 1) m )) m)))) (define (fermat-test n) (define a (+ 2 (random (- n 2)))) (= (expmod a n n) a)) Picking 47 randomly as a positive integer less than 97, the following function must return 47 if 97 is a prime according to Fermat’s little theorem. (remainder (fast-exp 47 97) 97) Now if we are using k to test n for being a prime in Fermat’s little theorem, if k is not returned then n can not be a prime. However, if the function does return k, it is very very likely that n is a prime, but not certain. So for example 77105 = 77 mod(105), even though 105 is not a prime. But trying more random k values adds evidence that n is prime. So 32105 = 92 mod(105), so 105 is not a prime. It would be worth while to test how often this false detection of a prime might occur, to get some estimate of the probability of such occurrence. For example 32
(fermat-test 9973) returns #t, (fermat-test 10000) returns #f Now this function fermat-test uses a random number generator supplied by scheme. This function is (random k) which supplies a integer in the interval [0, k] (maybe not including 0 or k, check this). However, some kind of pseudo random generator is being used by scheme, which can generate only a finite number of possibilities. So if k is larger than the number of possibilities, there is a problem. In that case use Fermat’s little theorem directly, and supply your own random number a to check for n being prime (expmod a n n) and check that the result is a. Now if the supplied a is greater than n, then if n is a prime, then the result b will not agree with a because the returned number will be in the interval [0, n − 1] while a is not. But of course b = a mod(n). See Fermat’s Theorem on p89 of Elementary Theory of Numbers, by Harriet Griffin, McGraw-Hill, 1954. If a and m are relatively prime then aφ(m) = 1mod(m), where φ is the Euler φ function, the number of integers less than m and relatively prime to m. If p is prime then φ(p) = p − 1, so ap−1 = 1mod(p) or ap = amod(p). This is Fermat’s little theorem.
16
A Python Program to Test For a Number Being Prime
# primetest.py 3/30/13 import random #+ function factorial def factorial(n): if(n == 0): return 1 33
return n*factorial(n-1) #+ function remainder of m divided by n, so m mod(n) def remainder(m,n): q=m/n return m - q*n #+ function square def square(n): return n*n #+ function fexp fast exponential function, that is, fast pow(m,n) def fexp(m,n): #print " entering fexp, n= ",n if(n == 0): return 1 if(remainder(n,2) == 0): # n is even #print " n is even " return square(fexp(m,n/2)) else: # n is odd #print " n is odd " return m*square(fexp(m,n/2))
#+ isprime Is an integer prime, using Fermat’s little theorem. def isprime(n,k): # test an integer n for being prime # repeat fermat’s test k times for, selecting a random a at each step. v=True for j in range(1,k): a = random.randint(2, n) m=fexp(a,n) r=remainder(m,n) # r= m mod(n) # a^n = a mod(n) if n is a prime. if (a!=r): 34
v=False break return v #program main print " d=a^n mod(n)\n" n1=1097 n2=1154 print " primes between ", n1," and ",n2 for n in range(n1,n2): v=isprime(n,10) if v: print n, " is prime" print "primes\n" print "101 103 107 109 113 127 131 137 139 149\n" print "151 157 163 167 173 179 181 191 193 197 199\n" print "881 883 887 907 911 919 929 937\n" print "941 947 953 967 971 977 983 991\n" print "997 1009 1013 1019 1021 1031 1033 1039\n" print "1049 1051 1061 1063 1069 1087 1091 1093\n" print "1097 1103 1109 1117 1123 1129 1151 1153\n" print "1163 1171 1181 1187 1193 1201 1213 1217\n"
17
Carmichael Numbers
Using the Python program of the previous section, I found a problem with the number 1105. The test always found this number to be a prime no matter how many values of a that I used. So finally I tested for all of the numbers with program carmichael1105.py It turns out that 1105 is a special number, called a Carmichael number or pseudoprime. This is a number such that the Fermat prime test fails for every number ”a.” 1105 is the second Charmichael number. That 561 is a Carmichael number can be seen with Korselt’s criterion. Indeed, n = 561 is
35
square free, and for each of the prime factors k of n, k − 1 divides n − 1. So 561 = 3 · 11 · 17 and 2|560, 10|560 and 16|560. The next six Carmichael numbers are (sequence A002997 in OEIS):
18
1105 = 5 · 13 · 17
(4 | 1104;
12 | 1104;
16 | 1104)
1729 = 7 · 13 · 19
(6 | 1728;
12 | 1728;
18 | 1728)
2465 = 5 · 17 · 29
(4 | 2464;
16 | 2464;
28 | 2464)
2821 = 7 · 13 · 31
(6 | 2820;
12 | 2820;
30 | 2820)
6601 = 7 · 23 · 41
(6 | 6600;
22 | 6600;
40 | 6600)
8911 = 7 · 19 · 67
(6 | 8910;
18 | 8910;
66 | 8910).
8911 = 7 · 19 · 67
(6 | 8910;
18 | 8910;
66 | 8910).
Fermat’s Last Theorem
We can find three integers x, y, z so that x2 + y 2 = z 2 . So for example 32 + 42 = 52 = 25 and 52 + 122 = 132 = 169. Fermat’s last theorem says that xn + y n = z n 36
has no solution in integers for n > 2 This has recently been proved by the English mathematician Andrew Wiles See Simon Singh’s documentary of Andrew Wiles’ extraordinary search for a proof: Youtube, Horizon: Fermat’s Last Theorem Taniyama-Shimura conjecture, elliptic curve of algebraic geometry modular forms, functions in the complex plane with elaborate symmetries.
19
Mersenne Primes
In modern mathematical usage, a Mersenne number is any number that is one less than a power of two, Mn = 2n − 1, where n is a positive integer greater than 1. They are named after the French monk Marin Mersenne (September 8,1588 to 1 September 1,1648), who studied them in the early 17th century. Mersenne numbers were in the past numbers of the form 2p − 1, where p is a prime. Mersenne primes are Mersenne numbers that are primes. The following numbers are Mersenne primes 22 − 1 = 3 = 11b 23 − 1 = 7 = 111b 25 − 1 = 31 = 11111b 27 − 1 = 127 = 1111111b. A Mersenne number is represented as a string of 1’s in binary. A necessary condition for 2n − 1 to be prime is for n itself to be prime. But this condition is not a sufficient condition. Although 11 is a prime number 211 − 1 = 2047 = 23 · 89.
37
To show this necessary condition, consider the homogeneous polynomial of degree n − 1 in the variables x and y, of the form. f (x, y) = xn−1 + xn−2 y + xn−3 y 2 + ... + y n−1. This polynomial is homogeneous because all terms are of the same degree, in this case degree n − 1. Multiplying by x − y we have (x − y)f (x, y) = (x − y)(xn−1 + xn−2 y + xn−3 y 2 + ... + y n−1) = (xn + xn−1 y + xn−2 y 2 + ... + xy n−1 ) −(xn−1 y + xn−2 y 2 + ... + xy n−1 + y n ) = xn − y n . So we obtain the identity xn − y n = (x − y)(xn−1 + xn−2 y + xn−3 y 2 + ... + y n−1). We can use this identity to show that if n is a composite number, then the Mersenne number 2n − 1, is not a prime. Proposition. The number 2n − 1 is not a prime if n is not a prime. Proof. Suppose n = jk is the product of two positive integers j and k, both greater than 1. Let x = 2j and y = 1 in the identity xk − y k = (x − y)(xk−1 + xk−2 y + xk−3 y 2 + ... + y k−1). Then xk = 2n , and y k = 1, so that 2n − 1 = (2j − 1)(2j(k−1) + 2j(k−2) + 2j(k−3) 1 + ... + 1), 38
is a product of two integers, each greater than 1, which proves the proposition. As of August 2008, only 44 Mersenne primes were known; the largest known prime number (232,582,657 − 1) is a Mersenne prime and in modern times the largest known prime has nearly always been a Mersenne prime. Like several previous Mersenne primes, it was discovered by a distributed computing project on the Internet, known as the Great Internet Mersenne Prime Search (GIMPS). GIMPS reported the finding of a potential 45th Mersenne prime on 2008-08-23, and a potential 46th on 2008-09-06. GIMPS now reports that two independent verifications of each potential M(p) prime confirm them as prime. An announcement about both primes will be made around 2008-09-17. Given the delay, chances are at least one of these has over 10 million decimal digits. From the identity (xn − y n ) = (x − y)(xn−1 + xn−2 y + +xn−3 y 2 + ... + y n−1), we can write 2mn − 1 = (2m − 1)(2m(n−1) + 2m(n−2) + 2m(n−3) + 1). It follows that if a Mersenne number is a prime, then it must be of the form 2p − 1, where p is a prime. There are 48 known Mersenne primes as of February 2013. n 1 2 3 4 5 6 7 8 9 17 48
p 2 3 5 7 13 17 19 31 61 2281 57,885,161
Mp 3 7 31 127 8191 131071 524287 2147483647 2305843009213693951 446087557...132836351 581887266...724285951
digits 1 1 2 3 4 6 6 10 19 687 17,425,170
Date 430BCE 430BCE 300BCE 300BCE 1456 1588 1588 1772 1883 Oct 1952 Jan 25,2013
Discoverer Greeks Greeks Greeks Greeks Anonomous Pietro Cataldi Pietro Cataldi Euler I. M. Pervushin Raphael Robinson Curtis Cooper
This last prime number is also the largest known prime number. 39
Method ... ... ... ... Trail Division Trial Division Trial Division Trial Division Lucus Sequences LLT SWAC GIMPS PC
20
The Chinese Remainder Theorem
Suppose we wish to compute an integer x that satisfies x = 2 mod(5) x = 3 mod(7) and x = 4 mod(11) The integers x − 2, x − 3, and x − 4 must simultaneously divisible by 5,7, and 11. Let K be the product of 5,7, and 11. Dividing K = 385 by each of 5,7, and 11 we get respectively K1 = 77, K2 = 55, K3 = 35. We have (77, 5) = 1, so 77 has an inverse in I5 , namely 3. We have (55, 7) = 1, so 55 has an inverse in I7 , namely 6. We have (35, 11) = 1, so 35 has an inverse in I11 , namely 6. Consider x = (2)(77)(3) + (3)(55)(6) + (4)(35)(6) = 2292. Now in mod( 5) the first term is 2, because (77)(3)=1, and the last two terms are zero, because 55 and 35 are multiples of 5. Similarly x is 3 mod (7), and x is 4 mod (11). The Chinese Remainder Theorem Let k1 , k2 , ..., kn be positive integers. If ki and kj are relatively prime for each i 6= j, the n equations x = a1 mod(k1 ) x = a2 mod(k2 ) ... x = an mod(kn ) have a unique integer solution. 40
Proof. Let K = k1 ...kn be the product of the ki0 s. For each ki define Ki = K/ki. Then Ki and ki are relatively prime, thus Ki has an inverse Ki0 in Iki . Write x = a1 K1 K10 + a2 K2 K20 + ... + an Kn Kn0 . In Ik1 x = a1 + 0... + 0 = a1 , because K1 K10 = 1 and k1 is a factor of the other terms. Thus x = a1 mod(k1 ). The other equations are satisfied by x similarly. To calculate Ki0 , calculate the gcd of Ki and k1 , namely, (Ki , ki) = 1, and in the process numbers so that 1 = αKi + βki. Then select a number Ki0 from the set {1, 2, 3, ..., ki − 1} that is congruent to α. See Apostol for applications of the Chinese Remainder Theorem.
21
Wilson’s Theorem
If p is a positive prime integer, then (p − 1)! + 1 = 0 mod p, or (p − 1)! + 1 is divisible by p. Equivalently (p − 1)! = −1 mod p. For example, if p = 7 then (p − 1)! + 1 = 721, and 721/7 = 103, so 721 is divisible by 7. From Wikipedia ”The theorem was first discovered by the Iraqi mathematician Ibn al-Haytham (known as Alhazen in Medieval Europe) circa 1000 AD, but it is named after John Wilson (a student of the English mathematician Edward Waring) who stated it in the 18th century.[1] Waring announced the theorem in 1770, although neither he nor Wilson could prove it. Lagrange gave the first proof in 1773. There is evidence that Leibniz was also aware of the result a century earlier, but he never published it.” 41
22
Hardy’s Cab Number 1729 = 123 + 13 = 93 + 103 .
23
Quadratic Residues
If x2 = a mod(m), has a solution, then a is called a quadratic residue of m.
24
Diophantine Equations
Diophantine (die oh fan tine) Equations are equations that are to be solved by integers.
25
Goldbach’s Conjecture
Goldbach’s conjecture is an open problem in mathematics. The conjecture is, that any even integer greater than 2, can be written as a sum of two primes. For example 4=2+2 6=3+3 8=3+5 10 = 3 + 7 12 = 7 + 5 ............ 24 = 11 + 13 ............ ............
42
(From Wikipedia) On 7 June 1742, the Prussian mathematician Christian Goldbach wrote a letter to Leonhard Euler (letter XLIII) [2] in which he proposed the following conjecture: Every integer greater than 2 can be written as the sum of three primes. He considered 1 to be a prime number, a convention subsequently abandoned. A modern version of Goldbach’s original conjecture is: Every integer greater than 5 can be written as the sum of three primes. Euler, becoming interested in the problem, replied by noting that this conjecture is equivalent to another version: Every even integer greater than 2 can be written as the sum of two primes, adding that he regarded this an entirely certain theorem (”ein ganz gewisses Theorema”), despite being unable to prove it.
26
The Riemann Zeta Function
Riemann’s zeta function is defined as the complex function ζ(s) =
∞ X
1 , s n=1 n
for the real part of s greater than 1. The definition is extended to the whole complex plane by analytic continuation. The Riemann zeta function is analytic everywhere except at s = 1 where it has a simple pole with residue 1. The Riemann zeta function has a role in the analytic proof of the prime number theorem. See Apostol.
27
The Riemann Hypothesis
In a memoir published in 1859 Riemann stated that the nontrivial zeroes of ζ(s) all lie on the vertical line in the complex plane that passes through (1/2,0). It has not yet been proved.
28
The Sieve of Eratosthenes
The Sieve of Eratosthenes is an algorithm for finding a list of prime numbers. It goes as follows. 43
Given a list of the first n positive integers, we can find all of the primes in the list by first crossing off all multiples of 2. Then the first number after 2 that is not crossed off is a prime number namely 3. Now we cross off all numbers in the list after 3, that are multiples of three. Then the first number in the list not yet crossed off is a prime, namely 5, then cross of all numbers in the list greater than 5 that are multiples of 5. If we continue this process until any multiple of the last prime is greater than n, then we have found all the primes in the list. Eratosthenes of Cyrene (276 to 194 BC), created this Algorithm to find all prime numbers less than n. Also Eratosthenes was the first to measure the circumference of the earth and was the head of the library at Alexandria. A computer algorithm is limited to the largest size integer that can be stored, commonly to the largest positive 32 bit integer. A sieve of Eratosthenes program: // sieve.cpp sieve of eratosthenes of Cyrene (276 to 194 BC) // Algorithm to find all prime numbers less than n. // Eratosthenes was the first to measure the // circumference of the earth. He was the head of the // library at Alexandria. // 3/20/2013 // 7/7/97 #include main(int argc,char **argv){ const int na=1000000; char a[na]; int i,k,n,j,m,n1,n2,ik; int prevp,twinp,twinp1,twinp2; //All indices of a, where a has value p (p: prime numbers), //are potential primes. //Computes all primes <= n if(argc < 2){ printf("Sieve of Eratosthenes of Cyrene (276 to 194 BC).\n"); printf("This is an Algorithm to find all prime numbers less than n.\n"); printf("Eratosthenes was the first to measure the circumference \n"); printf("of the earth. He was the head of the library at Alexandria.\n"); printf("\n"); 44
printf("sieve.cpp: James D. Emery, Version 3/20/13 \n"); printf("Print each of the prime integers p in the interval (n1,n2),\n"); printf("where 1 <= n1 < p < n2 <= 1,000,000 ,\n"); printf("and n1 and n2 are supplied on the command line. \n"); printf("If only one integer is supplied, we take it to be n2,\n"); printf("and set n1=1 by default. \n"); printf("Usage: sieve n1 n2 (or, sieve n2) \n"); exit(0); } if(argc > 2){ n1=atol(argv[1]); n2=atol(argv[2]); if((n1 < 1) || (n2 > na)){ cout << " Interval is out of range. "; exit(0); } } else{ n1=1; n2=atol(argv[1]); if(n2 > na){ cout << " Interval is out of range. "; exit(0); } }
// n=na-1; for(i=0;i
//set all multiples of k to c (c: composite numbers) for(i=2;i<=m;i++){ ik=i*k; if(ik
} } //Write a newline after 8 prime numbers have been printed. if(j%8 == 0){ cout << endl; } prevp=i;
} } cout << " \n Number of primes between " << n1 << " and " << n2 << " = " << j << cout << " \n Number of pairs of twin primes between " << n1 << " and " << n2 << cout << " \n The last pair of twin primes less than " << n2 << " = " << twinp1<< return 0; }
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Perfect Numbers
A perfect number is a positive integer equal to the sum of its positive divisors, so the first is 6, 6 = 1 + 2 + 3, the second is 28, 28 = 1 + 2 + 4 + 7 + 14 The next two are 496 and 8128.
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Twin Primes
Twin primes are prime numbers differing by two. It is not known if there are an infinite number of twin primes. This is a famous open problem. Here are a few twin primes: 3 5 5 7 17 19 29 31 47
41 43 59 61 71 73 101 103 107 109 137 139 149 151 179 181 191 193 227 229 281 283 311 313 347 349 419 421 431 433 461 463 521 523 569 571 599 601 617 619 641 643 809 811 827 829 857 859 881 883
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Analytic Number Theory
Analytic number theory is a subject that uses Real and Complex Analysis (Calculus) to prove number theoretic theorems such as the prime number theorem.
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Repeating Decimals
Using long division we know that any fraction or rational number has an infinite decimal representation with eventual repeating decimals. Let us consider the reverse problem of determining the fraction from the decimal expansion. We also discussed this above in the section called Decimal Expansions. We repeat here a similar analysis in a slightly less detailed form. Let a = .1234123412341234... have the four repeating decimals 1234. Show that a = 1234/9999. We have a − a/10000 = .1234 = 1234/10000 so a(1 − 1/10000) = 1234/10000 So multiplying by 10000 a(10000 − 1) = 1234, so a = 1234/9999. This can be generalized to the following. If a has n repeating decimals, d1 d2 d3 ....dn , that is a = .d1 d2 d3 ....dn d1 d2 d3 ....dn ..., then
d1 d2 d3 ....dn , 999...999 where the denominator has n digits, each a 9. a=
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33
Appendix, Primes Less Than 10000
2 3 19 23 53 59 89 97 131 173 223 263 311 359 409 457 503 569 613 659 719 769 827 881 941 997 1049 1097 1163 1223 1283 1321 1423 1459 1511 1571 1619 1693 1747 1811 1877 1949 2003 2069 2129 2203 2267 2311 2377 2423 2503 2579 2657 2693 2741 2801 2861
5 7 29 31 61 67 101 137 179 227 269 313 367 419 461 509 571 617 661 727 773 829 883 947 1009 1051 1103 1171 1229 1289 1327 1427 1471 1523 1579 1621 1697 1753 1823 1879 1951 2011 2081 2131 2207 2269 2333 2381 2437 2521 2591 2659 2699 2749 2803 2879
11 13 37 41 71 73 103 139 181 229 271 317 373 421 463 521 577 619 673 733 787 839 887 953 1013 1061 1109 1181 1231 1291 1361 1429 1481 1531 1583 1627 1699 1759 1831 1889 1973 2017 2083 2137 2213 2273 2339 2383 2441 2531 2593 2663 2707 2753 2819 2887
17 43 47 79 83 107 149 191 233 277 331 379 431 467 523 587 631 677 739 797 853 907 967 1019 1063 1117 1187 1237 1297 1367 1433 1483 1543 1597 1637 1709 1777 1847 1901 1979 2027 2087 2141 2221 2281 2341 2389 2447 2539 2609 2671 2711 2767 2833 2897
109 151 193 239 281 337 383 433 479 541 593 641 683 743 809 857 911 971 1021 1069 1123 1193 1249 1301 1373 1439 1487 1549 1601 1657 1721 1783 1861 1907 1987 2029 2089 2143 2237 2287 2347 2393 2459 2543 2617 2677 2713 2777 2837 2903
113 157 197 241 283 347 389 439 487 547 599 643 691 751 811 859 919 977 1031 1087 1129 1201 1259 1303 1381 1447 1489 1553 1607 1663 1723 1787 1867 1913 1993 2039 2099 2153 2239 2293 2351 2399 2467 2549 2621 2683 2719 2789 2843 2909
50
127 163 199 251 293 349 397 443 491 557 601 647 701 757 821 863 929 983 1033 1091 1151 1213 1277 1307 1399 1451 1493 1559 1609 1667 1733 1789 1871 1931 1997 2053 2111 2161 2243 2297 2357 2411 2473 2551 2633 2687 2729 2791 2851 2917
167 211 257 307 353 401 449 499 563 607 653 709 761 823 877 937 991 1039 1093 1153 1217 1279 1319 1409 1453 1499 1567 1613 1669 1741 1801 1873 1933 1999 2063 2113 2179 2251 2309 2371 2417 2477 2557 2647 2689 2731 2797 2857 2927
2939 3011 3079 3167 3221 3301 3347 3413 3491 3541 3607 3671 3727 3797 3863 3923 4003 4057 4129 4211 4259 4337 4409 4481 4547 4621 4673 4751 4813 4909 4967 5011 5087 5167 5233 5309 5399 5443 5507 5573 5653 5711 5791 5849 5897 6007 6073 6133 6211 6271 6329 6379 6473 6563 6637
2953 3019 3083 3169 3229 3307 3359 3433 3499 3547 3613 3673 3733 3803 3877 3929 4007 4073 4133 4217 4261 4339 4421 4483 4549 4637 4679 4759 4817 4919 4969 5021 5099 5171 5237 5323 5407 5449 5519 5581 5657 5717 5801 5851 5903 6011 6079 6143 6217 6277 6337 6389 6481 6569 6653
2957 3023 3089 3181 3251 3313 3361 3449 3511 3557 3617 3677 3739 3821 3881 3931 4013 4079 4139 4219 4271 4349 4423 4493 4561 4639 4691 4783 4831 4931 4973 5023 5101 5179 5261 5333 5413 5471 5521 5591 5659 5737 5807 5857 5923 6029 6089 6151 6221 6287 6343 6397 6491 6571 6659
2963 3037 3109 3187 3253 3319 3371 3457 3517 3559 3623 3691 3761 3823 3889 3943 4019 4091 4153 4229 4273 4357 4441 4507 4567 4643 4703 4787 4861 4933 4987 5039 5107 5189 5273 5347 5417 5477 5527 5623 5669 5741 5813 5861 5927 6037 6091 6163 6229 6299 6353 6421 6521 6577 6661
2969 3041 3119 3191 3257 3323 3373 3461 3527 3571 3631 3697 3767 3833 3907 3947 4021 4093 4157 4231 4283 4363 4447 4513 4583 4649 4721 4789 4871 4937 4993 5051 5113 5197 5279 5351 5419 5479 5531 5639 5683 5743 5821 5867 5939 6043 6101 6173 6247 6301 6359 6427 6529 6581 6673
2971 3049 3121 3203 3259 3329 3389 3463 3529 3581 3637 3701 3769 3847 3911 3967 4027 4099 4159 4241 4289 4373 4451 4517 4591 4651 4723 4793 4877 4943 4999 5059 5119 5209 5281 5381 5431 5483 5557 5641 5689 5749 5827 5869 5953 6047 6113 6197 6257 6311 6361 6449 6547 6599 6679
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2999 3061 3137 3209 3271 3331 3391 3467 3533 3583 3643 3709 3779 3851 3917 3989 4049 4111 4177 4243 4297 4391 4457 4519 4597 4657 4729 4799 4889 4951 5003 5077 5147 5227 5297 5387 5437 5501 5563 5647 5693 5779 5839 5879 5981 6053 6121 6199 6263 6317 6367 6451 6551 6607 6689
3001 3067 3163 3217 3299 3343 3407 3469 3539 3593 3659 3719 3793 3853 3919 4001 4051 4127 4201 4253 4327 4397 4463 4523 4603 4663 4733 4801 4903 4957 5009 5081 5153 5231 5303 5393 5441 5503 5569 5651 5701 5783 5843 5881 5987 6067 6131 6203 6269 6323 6373 6469 6553 6619 6691
6701 6779 6833 6907 6971 7027 7121 7207 7253 7349 7457 7517 7561 7621 7691 7757 7853 7919 8009 8087 8161 8231 8291 8369 8443 8537 8609 8677 8731 8803 8861 8941 9011 9091 9161 9227 9311 9377 9433 9491 9587 9649 9733 9791 9857 9929
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6703 6781 6841 6911 6977 7039 7127 7211 7283 7351 7459 7523 7573 7639 7699 7759 7867 7927 8011 8089 8167 8233 8293 8377 8447 8539 8623 8681 8737 8807 8863 8951 9013 9103 9173 9239 9319 9391 9437 9497 9601 9661 9739 9803 9859 9931
6709 6791 6857 6917 6983 7043 7129 7213 7297 7369 7477 7529 7577 7643 7703 7789 7873 7933 8017 8093 8171 8237 8297 8387 8461 8543 8627 8689 8741 8819 8867 8963 9029 9109 9181 9241 9323 9397 9439 9511 9613 9677 9743 9811 9871 9941
6719 6793 6863 6947 6991 7057 7151 7219 7307 7393 7481 7537 7583 7649 7717 7793 7877 7937 8039 8101 8179 8243 8311 8389 8467 8563 8629 8693 8747 8821 8887 8969 9041 9127 9187 9257 9337 9403 9461 9521 9619 9679 9749 9817 9883 9949
6733 6803 6869 6949 6997 7069 7159 7229 7309 7411 7487 7541 7589 7669 7723 7817 7879 7949 8053 8111 8191 8263 8317 8419 8501 8573 8641 8699 8753 8831 8893 8971 9043 9133 9199 9277 9341 9413 9463 9533 9623 9689 9767 9829 9887 9967
6737 6823 6871 6959 7001 7079 7177 7237 7321 7417 7489 7547 7591 7673 7727 7823 7883 7951 8059 8117 8209 8269 8329 8423 8513 8581 8647 8707 8761 8837 8923 8999 9049 9137 9203 9281 9343 9419 9467 9539 9629 9697 9769 9833 9901 9973
6761 6827 6883 6961 7013 7103 7187 7243 7331 7433 7499 7549 7603 7681 7741 7829 7901 7963 8069 8123 8219 8273 8353 8429 8521 8597 8663 8713 8779 8839 8929 9001 9059 9151 9209 9283 9349 9421 9473 9547 9631 9719 9781 9839 9907
6763 6829 6899 6967 7019 7109 7193 7247 7333 7451 7507 7559 7607 7687 7753 7841 7907 7993 8081 8147 8221 8287 8363 8431 8527 8599 8669 8719 8783 8849 8933 9007 9067 9157 9221 9293 9371 9431 9479 9551 9643 9721 9787 9851 9923
Fibonacci Numbers
See Jim Emery Fibonacci Numbers and the Golden Ratio, fibonacci.tex, fibonacci.pdf .
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Carl Fredich Gauss. The Prince of Mathematicians
(30 April 1777 23 February 1855) 1801: Disquisitiones Arithmeticae (Latin). A German translation by H. Maser Untersuchungen ber hhere Arithmetik (Disquisitiones Arithmeticae and other papers on number theory) (Second edition). New York: Chelsea. 1965. ISBN 0-8284-0191-8, pp. 1453. English translation by Arthur A. Clarke Disquisitiones Arithemeticae (Second, corrected edition). New York: Springer. 1986. ISBN 0-387-96254-9.
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Bibliography
[1]Apostol Tom, Introduction to Analytic Number Theory, SpringerVerlag, 1976. [2]Griffin Harriet, Elementary Theory of Numbers, McGraw-Hill, 1954. [3]Abelson, Harold and Sussman, Gerald J., Structure and Interpretation of Computer Programs, The MIT Press, McGraw-Hill, 1985. [4]Kranakis Evangelos, Primality and Cryptography, B. G. Teubner, Stuttgart, 1986,kck. [5]Gauss, Carl Friedrich, Disquisitiones Arithmeticae. Translated by Arthur A. Clarke, Yale University Press, 1966. (QA241 .G28 1870 ) (Gauss 1777-1855 1966).
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