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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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    @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

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Assuming you want positive integers $n$,

I believe we can show that

$(2n^2 + n)^2 \lt 4(n^4 + n^3 + n^2 + n + 1) \lt (2n^2 + n + 1)^2$

for $n \gt 3$.

Note: A similar inequality can be given for negative $n$.

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    @EricNaslund: Thanks! Corrected.2012-03-03
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Here is my answer!

H. Bensom, Germany

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    Your answer is very confusingly written and the special case you found ($n=3$) is already included in Byron Schmuland's comment.2014-09-05