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Fermat’s Little Theorem (2/24)

Fermat’s Little Theorem (2/24). Theorem ( flt ). If p is prime and GCD( a , p ) = 1, then a p – 1  1 (mod p ).

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Fermat’s Little Theorem (2/24)

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  1. Fermat’s Little Theorem (2/24) • Theorem (flt). If p is prime and GCD(a, p) = 1, then ap – 1 1 (mod p). • Again, this says that in a mod p congruence, we can reduce exponents by p – 1. Why? If n = k(p – 1) + r, thena n = ak(p – 1) + r= (a p– 1)k a r  1k a r = a r (mod p), where the congruence is by the theorem. • The proof of flt requires the following: • Lemma. If GCD(a, p) =1, then the set of numbers {a, 2a, 3a,..., (p – 1)a}, after all are reduced mod p, is just a rearrangement of the set of numbers {1, 2, 3,..., p – 1). • Example: Let p = 7 and let a = 3, then the set is{3, 6, 9, 12, 15, 18}  {3, 6, 2, 5, 1, 4} (mod 7)

  2. Proof of flt • a and p are as above. If we take the elements of the reduced set {a, 2a, 3a,..., (p – 1)a} and multiply them all together, we know we get 1 2  3 ... (p – 1) = (p – 1)!That is, a  2a 3a... (p – 1)a  (p – 1)! (mod p). • How many a’s are there here? • Factoring out the a’s, we get a p – 1 (p – 1)! (p – 1)! (mod p). • But finally, (p – 1)! is relatively prime to p (why?), so can cancel it!! • We arrive at a p – 1  1 (mod p). • 

  3. A Test for Compositeness • Fermat’s Little Theorem gives us a way to verify that a number is composite without factoring it! • Suppose n is some odd number and we’d like to know if it’s composite, but we’re having trouble factoring it. • Well, compute 2n –1 (mod n). What if the answer is not 1? • Example. I wonder if 376289 is prime? Using a computer, I find that 2376288  150132 (mod 376289). Conclusion? • In fact 376289 = 571 659, which is why I had trouble factoring it. • It turns out there are fast algorithms for computing powers to a modulus, but no known fast algorithms for factoring!

  4. But flt Is NOT “if and only if” • Unfortunately the converse of flt is not true, i.e., if GCD(a, n) = 1 and if an– 1  1 (mod n), we CANNOT conclude that n is prime! • The smallest counterexample with a base of 2 is 341. That is, 2340  1 (mod 341), BUT 341 is not prime (in fact, 341 = 11 31). Bummer! • 341 is called a 2-pseudoprime(i.e., “false prime with respect to base 2”). There are in fact infinitely many. • The smallest 3-pseudoprime is 91. Etc. • Really disturbing: A Carmichael number is a k-pseudoprime for every base k to which it is relatively prime. 561 is the smallest. There are infinitely many!!

  5. Assignment for Wednesday • Fully absorb these slides and all of Chapter 9. • We will not pursue pseudoprimes and Carmichael numbers further in this course, but if you’re interested, there are lots of things to study, including Chapter 19in our text. • Do Exercises 9.2 and 9.4.

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