BALKAN 1995 Problem 3

Let $a$ and $b$ be natural numbers with $a > b$ and having the same parity. Prove that the solutions of the equation $$x^2 - (a^2 - a + 1)(x - b^2 - 1) - (b^2 + 1)^2 = 0$$ are natural numbers, none of which is a perfect square.

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