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(2004). A (Not So) Complex Solution to a2 + b2 = cn. Mathematics Magazine: Vol. 77, No. 4, pp. 300-302.

One of the glorious chapters of elementary number theory, going back to Fermat and Euler, concerns the problem: which elements of the ring Z of rational integers are sums of two squares? From the point of view of arithmetic, the ring Z is truly the best of rings (however, algebraists often think Z the worst of rings, overwhelmingly devoid of structure). The tale I want to tell goes one step beyond elementary number theory by using the arithmetic of a second ring, the ring G of Gaussian integers, to solve the problem. I shall try to include enough background material and detail so that a motivated undergraduate could safely be unleashed upon this story, given an occasional assist in algebra.

(1986). Primitive Pythagorean Triples of Gaussian Integers. Mathematics Magazine: Vol. 59, No. 2, pp. 106-110.

1. As a matter of notation small case Latin letters will represent real integers and Greek letters will represent Gaussian integers. There are many representations for the complete residue system modulo n but the one that usually comes to mind is the set of integers {0, 1, 2, *..., n-1 } . Other representations are sometimes used; for example if (a, n) = 1 then {a, 2a, 3a, *, na } is a representation of the complete residue system modulo n. Another representation which has a somewhat pleasing quality is found in Uspensky and Heaslet [1] and is { x -n/2

Case 1. Suppose Ik Si is even. Since the lemma holds for m = k and k Si 1 ~~~~~~~~~~~~~~~~~~~~~~~1 is even, hence the expression sk + Sk1 +?s1 assumes every even integer from 1s to 0. It follows that the expression Sk+ 1 + Sk + Sk1 + ?+ s assumes every even (odd) integer from S1+j si to (Sk+?1 Si), if 1 Si is even (odd). By assumption, (Sk+11 si) is an even integer less than 1, if Sk+1 is even. If Sk+1 is odd, then (Sk+1X ks) < 1.

The purpose of this paper is to demonstrate the usefulness of the binary quadratic form, (1) x2 + pqy2 in integers x and y with integral coefficients, in the factorization of odd integers of the form 4k + 1, where k is a positive integer. We begin by introducing the following two lemmas. Lemma 1. If Al is a positive integer of the form 4k + 1 and r1 and r2 are any pair of positive integral roots of M where M = r1r2, there existintegers u, m, v, n, p, and q such that: (i) r=pu 2+qv2 (ii) r2= pm2 + qn2 Proof: Let p = v = n = 1