(Inspired by IMO 1972 Longlist 1, treated)

Find all integers $x$ such that $x^{4}+x^{3}+x^{2}+x+1$ is a perfect square.
Suppose $x$ is a solution, then $4\left(x^{4}+x^{3}+x^{2}+x+1\right)$ is also a perfect square. Now, $\left(2 x^{2}+x\right)^{2}=4 x^{4}+4 x^{3}+x^{2}, 4\left(x^{4}+x^{3}+x^{2}+x+1\right)>\left(2 x^{2}+x\right)^{2}$ if and only if $3 x^{2}+4 x+4>0$, if and only if $2 x^{2}+(x+2)^{2}>0$, which is always true.

Furthermore, $4\left(x^{4}+x^{3}+x^{2}+x+1\right)0$, if and only if $(x-3)(x+1)>0$, which is true for $x>3$.

Therefore, the only potential solutions are for $-1 \leq x \leq 3$. We test:

| $x$ | $x^{4}+x^{3}+x^{2}+x+1$ | Perfect square? |
| :---: | :---: | :---: |
| -1 | $1=1^{2}$ | YES |
| 0 | $1=1^{2}$ | YES |
| 1 | 5 | no |
| 2 | 31 | no |
| 3 | $121=11^{2}$ | YES |

In conclusion, there are also three solutions, $x=-1, x=0$ and $x=3$.