Under what condition on $(x, y) \in \mathbb{N}^{2}$ is the integer $(x+y)^{2}+3 x+y+1$ a perfect square?
For $x, y \in \mathbb{N}$ we notice that:

$$
(x+y)^{2}<(x+y)^{2}+3 x+y+1<(x+y)^{2}+4(x+y)+4=(x+y+2)^{2}
$$

Therefore, $(x+y)^{2}+3 x+y+1$ is a perfect square if and only if $(x+y)^{2}+3 x+y+1=(x+y+1)^{2}$. We solve the equation:

$(x+y)^{2}+3 x+y+1=(x+y+1)^{2} \Longleftrightarrow(x+y)^{2}+3 x+y+1=(x+y)^{2}+2(x+y)+1 \Longleftrightarrow x=y$

The required condition is therefore $x=y$.