Applied abstract algebra (draft) by David Joyner, Richard Kreminski, Joann Turisco

By David Joyner, Richard Kreminski, Joann Turisco

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Consider the kth row: k m + k 2m + k (n - 1)m + k , where gcd(k , m) = 1 . Suppose im + k = jm + k ( mod n) , with 0 :::; i, j < n. Then n divides im + k - (jm + k) = (i - j)m. But, since gcd(m , n) = 1, it follows from Proposition 1 . 2 . e. i = j ( mod n) . This, together with the above inequalities involving i and j show that i = j . Therefore, no two elements in the kth row are congruent ( mod n) . 9, every element in the kth row is relatively prime to m. Since no two elements are congruent, their least residues are 0, 1 , 2 , .

Aa "' 2 "' ( mod m) . 1 . 7. CONGRUENCES 55 Example 1 . 7. 10. We compute 10 11 (mod 21) . First, 11 = 8 + 2 + 1, so we compute • • • • 10 ( mod 2 1 ) = 10, 1 02 ( mod 21) = 16, 1 04 ( mod 21) = 1 62 = 4, 1 08 ( mod 21) = 42 = 16. Thus 10 11 ( mod 21) = 4 · 16 · 10 = 19. 1 . 7. 5 Fermat 's little theorem We have seen that if p is a prime then p divides all the binomial coefficients ( : ), 1 ::::; k ::::; p - 1 . Because of this and the binomial theorem, we have (a + b)P = aP + lf (mod p) , for any integers a, b.

X = ak ( mod nk ) ( 1 . 7) has a simultaneous solution x E Z . Furthermore, if x, x ' are two solutions to (1. 7} then x' = x (mod n) . This follows from the k = 2 case proven above using mathematical induc­ tion. The details are left as an exercise. We give a different proof below. proof: As a runs over all n integers 0 ::::; a < n, the k-tuples (a ( mod n1 ) , . . , a ( mod nk )) form a collection of n distinct k-tuples in { 0, 1 , . . , n - 1 } k . ( Exercise: show why they are distinct.

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