By David Joyner, Richard Kreminski, Joann Turisco

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9. Let a > 0, b > 0 and m and only if there is an integer x such that ax > 1 be integers. gcd(a, m ) l b if b ( mod m) . The result above tells us exactly when we can solve the "modulo m analogs" of the equation ax = b studied in elementary school. The proof (which requires the previous lemma and Proposition 1 . 2 . 16) is left as a good exercise. 1 . 4 Repeated squaring algorithm How hard do you think it would be to compute by hand 2 1 28 mod 5? If you guess too hard, you're wrong. The algorithm described below shows how such seemingly difficult calculations are actually quite easy.

Ao + ad O + a2 100 - a3 - a4 10 - a5100 - a6 103 - a 7 104 - . . (mod 7) = ao + ad O + a2 100 - a3 - a4 10 - a5 100 + a6 + a 7 10 + . . (mod 7) from which the divisibilty result for 7 follows. We have 1000 = 8 125, so 1000 = 0 ( mod 8) . Substituting, we obtain · ao + a 1 10 + a2 100 + a3 1000 + a4 104 + a5105 + a6 106 + a7 10 7 + . . = ao + a 1 10 + a2 100 ( mod 8) from which the divisibilty result for 8 follows. The following result, which will be used later, is a consequence of the Euclidean algorithm.

This x exists and is unique mod mn, by the Chinese remainder theorem, so f is well-defined - provided we show x E * mn · To show this, we must show that (x, mn) = 1 if (x , m) = (x, n) = 1 , and that if (a, m) = 1 and x = a (mod m) , then (x, m) = 1 . We leave these as exercises. f is 1-1 : If x = f (a, b) = f (a', b') then x = a ( mod m) , x = b ( mod n) and x = a' ( mod m) , x = b' (mod n) . *