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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. –  Henning Makholm Mar 3 '12 at 19:43
5  
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 $|x|>1$. –  Byron Schmuland Mar 3 '12 at 20:43
    
Wow! that was very useful to know –  Sam Mar 3 '12 at 21:39
    
$n^4 + 4 n^3 + 6 n^2 + 4 n + 1$ works better. –  Will Jagy Mar 3 '12 at 21:49
    
which is $(n+1)^4 = \left(\left(n+1\right)^2\right)^2$ –  Sam Mar 3 '12 at 23:08

2 Answers 2

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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Yes $n$ is positive integer. –  Sam Mar 3 '12 at 19:46
2  
Since you need $n^2>2n+3$, you must have $n>3$. Note in particular, when $n=3$, $n^4+n^3+n^2+n+1=121=11^2$. –  Eric Naslund Mar 3 '12 at 19:53
    
Yes I got it Aryabhata. That was a fantastic hint. –  Sam Mar 3 '12 at 19:56
    
I did get for n=3, but proving that it cannot have any other $n$ is where I can use Aryabhata's hint. –  Sam Mar 3 '12 at 19:56
    
Thanks Eric and Aryabhata –  Sam Mar 3 '12 at 20:00

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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. –  Joonas Ilmavirta Sep 5 at 13:51

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