Competition problem — Number theory

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Number theory Difficulty 4.9 AIME Find the answer

Find the smallest possible value of $x+y$ where $x, y \geq 1$ and $x$ and $y$ are integers that satisfy $x^{2}-29y^{2}=1$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Continued fraction convergents to $\sqrt{29}$ are $5, \frac{11}{2}, \frac{16}{3}, \frac{27}{5}, \frac{70}{13}$ and you get $70^{2}-29 \cdot 13^{2}=-1$ so since $(70+13\sqrt{29})^{2}=9801+1820\sqrt{29}$ the answer is $9801+1820=11621$

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.