1. Complexity1 Pratt’s Theorem Proved

2. Complexity2 Introduction So far, we’ve reduced proving PRIMES  NP to proving a number theory claim. This is our next task.

3. Complexity3 Main Theorem Claim: A number p>2 is prime iff there exists a number 1<r<p (called primitive root) s.t 1) r p-1 = 1 (mod p) 2)  prime divisor q of p-1: r (p-1)/q  1 (mod p) PAP 222-227

4. Complexity4 Euler’s Function  (n) = { m : 1  m<n  gcd(m,n)=1 } Euler’s function:  (n)=|  (n)|.  (12)={1,5,7,11}  (12)=4 Example: Observe: For any prime p,  (p)={1,...,p-1}

5. Complexity5 An Equivalent Definition of Euler’s Function Using Prime Divisors Let p be a prime divisor of n. The probability p divides a candidate is 1/p. Thus:... all the residues modulo n are candidates for  (n)

6. Complexity6 Corollaries Corollary: If gcd(m,n)=1,  (mn)=  (m)  (n). Proof:  (6)=|{1,5}|=2  (2)=|{1}|=1  (3)=|{1,2}|=2  (6)=|{1,5}|=2  (2)=|{1}|=1  (3)=|{1,2}|=2

7. Complexity7 21=7·3  (21)={1,2,4,5,8,10,11,13,16,17,19,20}  (3) ={1,2}  (7) ={1,2,3,4,5,6} 21=7·3  (21)={1,2,4,5,8,10,11,13,16,17,19,20}  (3) ={1,2}  (7) ={1,2,3,4,5,6} Corollaries The Chinese Remainder Theorem: If n is the product of distinct primes p 1,...,p k, for each k-tuple of residues (r 1,...,r k ), where r i  (p i ), there is a unique r  (n), where r i =r mod p i for every 1  i  k.

8. Complexity8 The Chinese Remainder Theorem Proof: If n is the product of distinct primes p 1,...,p k, then  (n)=  1  i  k (p i -1). This means |  (n)|=|  (p 1 ) ...  (p k )|. The following is a 1-1 correspondence between the two sets: r (r mod p 1,...,r mod p k )

9. Complexity9 Another Property of the Euler Function Claim:  m|n  (m)=n.  m|12  (m)=  (1) +  (2) +  (3) +  (4) +  (6) +  (12)= |{1}| + |{1}| + |{1,2}| + |{1,3}| + |{1,5}| + |{1,5,7,11}|= 1 + 1 + 2 + 2 + 2 + 4 = 12  m|12  (m)=  (1) +  (2) +  (3) +  (4) +  (6) +  (12)= |{1}| + + |{1,2}| + |{1,3}| + |{1,5}| + |{1,5,7,11}|= 1 + 1 + 2 + 2 + 2 + 4 = 12 Example:

10. Complexity10 Another Property of the Euler Function Claim:  m|n  (m)=n. Proof: Let  1  i  l p i k i be the prime factorization of n.  (n)=n  p|n (1-1/p) telescopic sum  m|n  (m)= Since  ( ab )=  ( a )  ( b )  ( ab )=  ( a )  ( b )

11. Complexity11 Fermat’s Theorem Fermat’s Theorem: Let p be a prime number.  0<a<p, a p-1 mod p=1 p=5; a=2 2 5-1 mod 5 = 16 mod 5 = 1 p=5; a=2 2 5-1 mod 5 = 16 mod 5 = 1 Example:

12. Complexity12 Observation  0<a<p, a·  (p):={a·m (mod p) | m  (p)} =  (p) 2413 1243  (5) ·2 (mod 5) Example:

13. Complexity13 Fermat’s Theorem: Proof Therefore, for any 0<a<p:  0 (mod p)

14. Complexity14 Generalization Claim: For all a  (n), a  (n) =1 (mod n). n=8,  (8) = {1,3,5,7} 3 4 =1 (mod 8) n=8,  (8) = {1,3,5,7} 3 4 =1 (mod 8) Example:

15. Complexity15 Generalization: Proof 1357 1375  (8) Example: 1 3 7 5 * (mod 8) 3157 5 7 713 351 Again: For any a  (n), a·  (n)=  (n) Again:  m  (n)  0 (mod n) And the claim follows.

16. Complexity16 Exponents Definition: If m  (p), the exponent of m is the least integer k>0 such that m k =1 (mod p). p=7, m=4  (7), the exponent of 4 is 3. p=7, m=4  (7), the exponent of 4 is 3. Example:

17. Complexity17 All Residues Have Exponents Let s  (p).  j>i  N which satisfy s i =s j (mod p). s i is indivisible by p.  s j-i =1 (mod p).

18. Complexity18 Observe the equation x k =1 (mod p). Assume s is a solution whose exponent is k. Then 1,s,...,s k-1 are distinct. (Just as in...)Just as in... Note they are all solutions of the equation. In fact, there are no other solutions, as implied by the following claim.

19. Complexity19 Polynomials Have Few Roots Claim: Any polynomial of degree k that is not identically zero has at most k distinct roots modulo p. Proof: By induction on k. Trivially holds for k=0. Suppose it also holds for some k-1. By way of contradiction, assume x 1,...,x k+1 are roots of  (x)=a k x k +...+a 0.  ’(x)=  (x)-a k  1  i  k (x-x i ) is of degree k-1 and not identically zero. x 1,...,x k are its roots - Contradiction!

20. Complexity20 Summary We’ve completed Pratt’s theorem proof, stating PRIMES  NP  coNP. 