for which values of $n$, $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square?

Question

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.

Can't do it using modular arithmetic, because at $n=0$ the expression is a perfect square modulo all $n$. — Sarvesh Ravichandran Iyer, Sep 12 '20 at 11:48
Suppose we modulo $p$. Then $n=kp$ cannot produce a contradiction. This is what the FBI agent meant. — player3236, Sep 12 '20 at 11:52
Possible duplicate of Diophantine equation with quartic polynomial — Parcly Taxel, Sep 12 '20 at 12:01
Or does this answer your question? What are integers $x$ and $n$ such as $ x^4+x^3+x^2+x+1 = n^2$? — Dietrich Burde, Sep 12 '20 at 12:01

score 3 · Accepted Answer · answered Sep 12 '20 at 12:15

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 $(x,y)=(3,11),(0,1)\space\space\text{and}\space\space(-1,1).$

for which values of $n$, $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square?

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