By Andrew Granville

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**Extra resources for An Introduction to Gauss's Number Theory**

**Example text**

2. Let f (x) ∈ Z[x]. Suppose that f (r) ≡ 0 (mod m) for all integers r in the range 0 ≤ r ≤ m − 1. Deduce that there does not exist an integer n for which f (n) = 0. 3 to determine all of the squares modulo m, for m = 3, 4, 5, 6, 7, 8, 9 and 10. ) (b) Show that there are no solutions in integers x, y, z to x2 + y 2 = z 2 with x and y odd. ) (c) Show that there are no solutions in integers x, y, z to x2 + y 2 = 3z 2 with (x, y) = 1. (d) Show that there are no solutions in integers x, y, z to x2 + y 2 = 6z 2 with (x, y) = 1.

2. Abstractions. 1, a result whose formulation is not obviously relevant, and yet was used to good effect. The ancient Greek mathematicians recognized that abstract lemmas allowed them to prove sophisticated theorems. The archetypal result is “Euclid’s Lemma”, an important result that first appeared in Euclid’s “Elements” (Book VII, No. 1: Euclid’s Lemma. If c divides ab and gcd(c, a) = 1 then c divides b. Proof of Euclid’s lemma. 3. Now c divides both c and ab, so that c divides c · bm + ab · n = b(cm + an) = b.

Mk ] such that x ≡ aj (mod mj ) for each j ? One idea is to re-write each congruence x ≡ aj (mod mj ) as the set of congruences x ≡ aj e (mod p1j,1 ), x ≡ aj e where p1 , . . , pr are the distinct primes dividing m, and mj = we have the set of congruences x ≡ a1 e (mod pi 1,i ), x ≡ a2 (mod perj,r ), (mod p2j,2 ), . . , x ≡ aj r i=1 e (mod pi 2,i ), . . , x ≡ ak e pi j,i . Now for each i, e (mod pi k,i ), and evidently, if ei := max{ej,i : 1 ≤ j ≤ k} = ej(i),i , then there exists such an x if and e only if aj ≡ aj(i) (mod pi j,i ) for each j.

### An Introduction to Gauss's Number Theory by Andrew Granville

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