Olympiad Maths Prep

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Problem 331

AMC 12 late, AIME early
Number theory Difficulty 4.7 Prove it IRL_ABooklet_2020 · Ireland

The triple (1,5,7)(1, 5, 7) is such that the squares (1,25,49)(1, 25, 49) are in arithmetic progression. Show that there are infinitely many triples (a,b,c)(a, b, c) of relatively prime positive integers such that a2a^2, b2b^2 and c2c^2 are in arithmetic progression.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let b2=u2+v2b^2 = u^2 + v^2 be a Pythagorean triple with integers u<v<bu < v < b. Then
(v−u)2+(v+u)2=v2−2uv+u2+v2+2uv+u2=2u2+2v2=2b2. (v-u)^2 + (v+u)^2 = v^2 - 2uv + u^2 + v^2 + 2uv + u^2 = 2u^2 + 2v^2 = 2b^2.
Rearranging,
b2−(v−u)2=(v+u)2−b2. b^2 - (v-u)^2 = (v+u)^2 - b^2.
Thus, setting (a,b,c)=(v−u,b,v+u)(a, b, c) = (v-u, b, v+u) produces squares in arithmetic progression. It is clear that no two Pythagorean triples produce the same arithmetic progression, and that relative primality is preserved. Thus, the infinitude of Pythagorean triples implies there are infinitely many square arithmetic progressions.

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