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I tried every $n \in [4;2,000,000]$ and it seems to be true that there will be no other square numbers.

Can someone think of a proof for this conjecture?

Any kind of help will be appreciated!

2 Answers2

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No. Assume $$n^4+n^3+n^2+n+1=\Bigl(n^2+\frac n2+x\Bigr)^2,$$ where $x$ is a half integer (integer multiple of $\frac12$).

Eapanding, you find that $x=0$ is too small and $x=1$ is too large, so $x=\frac12$ is the only possibility. Substitute that in, and get a quadratic equation to determine $n$. The only positive solution is $n=3$.

(How I thought of this: The square root of the left hand side must be close to $n^2$, and on second thought much closer to $n^2+n/2$. Let $x$ be the difference.)

Harald Hanche-Olsen
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  • Thank you very much for your answer. But I still can't understand why $x=0$ is too small and $x=1$ too large. Would you care to elaborate? –  Mar 07 '15 at 17:40
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    @Silenttiffy Try $x=0$: $(n^2+n/2)^2=n^4+n^3+n^2/4<n^4+n^3+n^2+n+1$. Or try $x=1$: $(n^2+n/2+1)^2=n^4+n^3+\frac94n^2+n+1>n^4+n^3+n^2+n+1$. – Harald Hanche-Olsen Mar 07 '15 at 20:54
  • Oh, got it! Thank you. And why do we now that if $x=0$ is too small and $x=1$ too large that $x$ has to be equal to $x=0.5$? –  Mar 08 '15 at 15:53
  • @Silenttiffy Because we're looking for a perfect square, meaning the square of an integer, so $n^2+n/2+x$ has to be an integer. Given that $n$ is an integer, that means $2x$ must be an integer. – Harald Hanche-Olsen Mar 09 '15 at 10:17
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The solution is given in Ribenboim's book on Catalan's conjecture, where all Diophantine equations $$y^2=1+x+x^2+\cdots +x^k$$ are studied. For $k=4$, the only positive solution is $x=3$.

Dietrich Burde
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