Prove that if a²+b² is a multiple of ab+1, for positive integer a and b, then (a²+b²)/(ab+1) is a perfect square.

Let (a, b) be any solution pair with a>b. Let s = (aČ+bČ)/(ab+1). Then another solution can be derived by creating solution pair (s*a-b, a).

Plugging in the trivial solution of (a, 0) with a>1 and repeatedly applying the transform above creates the following sequence:

(a, 0)

(a^3, a)

(a^5-a, a^3)

(a^7-2a^3, a^5-a)

(a^9-3a^5+a, a^7-2a^3)

(a^11-4a^7+3a^3, a^9-3a^5+a)

(a^13-5a^9+6a^5-a, a^11-4a^7+3a^3)

(a^15-6a^11+10a^7-4a^3, a^13-5a^9+6a^5-a)

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For different choices of a, I believe this algorithm will generate every solution.