3. Find all integers $x, y$ for which $x+y, 2x+3y$ and $3x+y$ are perfect squares.
Answer: $x=y=0$. Let $x+y=a^{2}, 2 x+3 y=b^{2}$ and $3 x+y=c^{2}$. Note that $2 b^{2}+c^{2}=7 a^{2}$. We will prove that this equation has a unique solution: $a=b=c=0$. Indeed, let $(a, b, c)$ be a solution with the minimal sum $a^{2}+b^{2}+c^{2}$, in which not all numbers $a, b, c$ are equal to 0. Then $2 b^{2}+c^{2}: 7$. It is easy to see that this is possible only if $b \vdots 7$ and $c \vdots 7$. But then $7 a^{2} \vdots 49$, from which $a \vdots 7$. Reducing $a, b, c$ by 7, we get a new non-zero solution with a smaller sum $a^{2}+b^{2}+c^{2}$. Contradiction. Thus, $a=b=c=0$. Then $x=\frac{c^{2}-a^{2}}{2}=0$ and $y=a^{2}-x=0$.