Problem 2. Solve, in the set of integers, the equation:

$$
2 x^{2}+2 y^{2}-3 x-3 y+2=0
$$

Constantin Apostol, Gazeta Matematică nr. 9/2014
## Solution and marking scheme

Multiplying by 2, the equality from the statement is equivalently written as $(x-3)^{2}+(y-3)^{2}+3 x^{2}+3 y^{2}=14$

Then $3 x^{2} \leq 14$, from which $|x| \leq 2$, that is $x \in\{-2,-1,0,1,2\}$

By verifying each case separately, a single solution is obtained, namely $x=y=1$