KöMaL (Metresis) 1894, Problem 12

Problem:
Find a four digit number which is a perfect square, knowing that the first two digit number exceeds by $1$ the last two digit number.

Solution:
Let $x$ be the first two digit number and let $y^2$ be the four digit number. Then, $31 < y < 100$ and
$$100(x+1)+x=y^2 \iff 101x=(y-10)(y+10).$$ So, $101$ divides one between $y-10$ and $y+10$. But $y-10<101$, so $101|(y+10)$. Since $41<y+10<110$, $y+10=101$, which gives $y=91$. Therefore, $y^2=8281$.