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Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?

I tried to simplify

\begin{align*} n^4+n^3+n^2+n+1 &= n^2(n^2+1)+n(n^2+1)+1\\ &= (n^2+n)(n^2+1)+1 \\ &= n(n+1)(n^2+1)+1 \end{align*} Then I assumed that the above expression is a square; then

$$ n(n+1)(n^2+1)+1 = k^2$$

$$ \begin{align*} n(n+1)(n^2+1) &= (k^2-1) \\ &= (k+1)(k-1) \end{align*} $$

Then trying to reason with prime factors, but cannot find a concrete proof yet.

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    $n=0$ and $n=-1$ both work.2012-03-03
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    Ribenboim's book on Catalan's conjecture has a detailed analysis of the Diophantine equation $v^2=1+x+x^2+\cdots+x^{n−1}$. The only non-trivial solutions are $n=5,x=3$ and $n=4,x=7$. By non-trivial, I mean $|x|>1$.2012-03-03
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    Wow! that was very useful to know2012-03-03
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    $n^4 + 4 n^3 + 6 n^2 + 4 n + 1$ works better.2012-03-03
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    which is $(n+1)^4 = \left(\left(n+1\right)^2\right)^2$2012-03-03
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    Alright, if you insist.2012-03-03
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    @ByronSchmuland Is that book available for a reasonable price anywhere? If not, do you know if it's [legally] available on line? A Google search seems to indicate not.2014-09-05
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    @rogerl Not that I know of. We have a copy in our university library, which is how I got ahold of it.2014-09-05

2 Answers 2