for which values of $n$, $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square? I saw a question my number theory notes:
for which values of $n$ , $1+n+n^{2}+n^{3}+n^{4}$ is a perfect  square where $"n"$ is non negative integer?
It was solved my professor using inequalities,but i think that his method is too long.Because of that, i did not write it here.
I tried to solve it using modular arithmetic but i could not do it.
I am looking for hints or solutions to solve it using modular artihmatic.
Note= I tried to use $(mod4)$ because for all $x^{2} \equiv \pm 1 (mod4)$ but i stuck in it.

Following from the comments.
We want to find integers $(x,y)$ such that $x^4+x^3+x^2+x+1=y^2.$ Clearly if $x^4+x^3+x^2+x+1$ is a square then so is $f(x)=4x^4+4x^3+4x^2+4x+4.$
Note that $f(x)> (2x^2+x)^2$ $\forall x\in\mathbb R$ and also $f(x)=4x^4+4x^3+4x^2+4x+4=(2x^2+x+1)^2-(x+1)(x-3).$ Since the last term is non-positive in the interval $[-1,3]$  it follows that
$$(2x^2+x)^2<f(x)<(2x^2+x+1)^2$$
for all $x\notin[-1,3].$ So $f(x)$ cannot itself be a square  for $x\notin[-1,3]$ since it is bounded by two consecutive squares.
Thus we only need to check for integers in $[-1,3]$ and we find the solutions $(x,y)=(3,11),(0,1)\space\space\text{and}\space\space(-1,1).$
