HMMT-Feb 2017 guts, Problem 27

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

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 $9801+1820=11621$ $\fbox{11621}$.

Answer: 11621