(treatise)

For what condition on $y \in \mathbb{N}$ is $y^{2}+5 y+12$ a square?
We know that $(y+2)^{2}=y^{2}+4 y+4y^{2}+5 y+12$ if and only if $6 y+9>5 y+12$, if and only if $y>3$.

Therefore, for $y \geq 4, (y+2)^{2}<y^{2}+5 y+12<(y+3)^{2}$, under these conditions $y^{2}+5 y+12$ cannot be a square.

Finally, the only potential solutions are $0 \leq y \leq 3$. We test:

| $y$ | $y^{2}+5 y+12$ | Perfect square? |
| :---: | :---: | :---: |
| 0 | 12 | no |
| 1 | 18 | no |
| 2 | 26 | no |
| 3 | $36=6^{2}$ | YES |

In the end, there is a unique solution $y=3$.