2011 PuMAC Problems/NT Problem A8

By multiplying both sides by 2 and simplifying, the condition becomes $(2a+1)^2 - 2b^2 = -1$ . This is a variant of Pell's Equation $x^2 - 2y^2 = -1$ with minimal solution $(x_0, y_0) = (1,1)$ . We can obtain all solutions $(x_0, y_0), (x_1, y_1), \ldots$ by computing $x_k + y_k\sqrt{2} = (1 + \sqrt{2})^{2k+1}.$

In particular, we get $(x_1, y_1) = (7, 5)$ , $(x_2, y_2) = (41, 29)$ , and $(x_3, y_3) = (239, 169)$ . This third solution corresponds to $a = \boxed{119}$ . Indeed, $119, 120, 169$ is the third consecutive Pythagorean triple.

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2011 PuMAC Problems/NT Problem A8