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Number theory Difficulty 4.7 AIME Prove it 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.

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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