2011 PuMAC Problems/NT Problem A8

Revision as of 15:30, 22 June 2020 by Pureswag (talk | contribs) (Created page with "By multiplying both sides by 2 and simplifying, the condition becomes <math>(2a+1)^2 - 2b^2 = -1</math>. This is a variant of Pell's Equation <math>x^2 - 2y^2 = -1</math> with...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.