Find all odd integers $y$ such that there exists an integer $x$ satisfying

$$
x^{2}+2 y^{2}=y x^{2}+y+1
$$

## 2 Solution
We rewrite the equation in the form

$$
(y-1) x^{2}=2 y^{2}-y-1
$$

since $2 y^{2}-y-1=(y-1)(2 y+1)$, therefore if $y \neq 1$, we can divide by $y-1$, and the equation becomes $x^{2}=2 y+1$. Since $y$ is odd, it can be written in the form $y=2 k+1$. In this case, $x^{2}=4 k+3$, but a table of congruences shows that $x^{2}$ can only be 0 or 1 modulo 4, and thus cannot be of the form $4 k+3$. It remains to handle the case $y=1$, and we easily notice that it leads to a solution for all $x$. The only integer $y$ that is a solution to this problem is therefore $y=1$.

## 2 Group B

## 1 Statement