Let $n,k$ two integers greater than $1$, is it possible that $n(n+1)(n+2)...(n+k)$ is a square $m^2$, with $m$ an integer ?

Thanks in advance.

  • $\begingroup$ Isn't this reducible via completing the square to Pell's equation in the case where $k=1$? $\endgroup$ – Michael Hardy Sep 21 '11 at 18:11
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    $\begingroup$ @Michael $n(n+1)$ is never a square for $n > 0$. Because $n$ and $n+1$ are relatively prime, the only way for $n(n+1)$ is a square is to that each of $n$ and $n+1$ is individually a perfect square. It is easy to see that this is impossible unless $n=0$. $\endgroup$ – Srivatsan Sep 21 '11 at 18:14
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    $\begingroup$ (I'll record another argument for $k=1$. Hopefully someone can see a proof for general $k$. :)) The number $n(n+1)$ is strictly between $n^2$ and $(n+1)^2$, and hence is not a perfect square. $\endgroup$ – Srivatsan Sep 21 '11 at 18:20
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    $\begingroup$ I see. According to Wikipedia: <b>Pell's equation<b> is any Diophantine equation of the form $$x^2-ny^2=1$$ where $n$ is a nonsquare integer. I missed the "nonsquare" part. $\endgroup$ – Michael Hardy Sep 21 '11 at 19:45
  • $\begingroup$ cant you just use bertrand postulate an the fact that the largest prime less than n+k divides the product? $\endgroup$ – Robert William Hanks Sep 22 '11 at 11:45

The answer is no, it can never be a square. This problem was originally solved by Erdos in 1939. The paper can be found here.

Later, in 1975 Erdos and Selfridge improved the result and solved a longstanding conjecture which was first considered by Liouville in the 19th century, by showing that the product of two or more consecutive positive integers is never a perfect power.

  • $\begingroup$ Thanks for your answer, Eric. $\endgroup$ – francis-jamet Sep 21 '11 at 18:33
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    $\begingroup$ Here's the link to Erdos 1975 paper: renyi.hu/~p_erdos/1975-46.pdf Some other papers from renyi.hu/~p_erdos/Erdos.html might be of interest too. $\endgroup$ – Martin Sleziak Sep 21 '11 at 18:37
  • $\begingroup$ @Jason Can you explain your final comment? What convinced you that the problem is difficult for perfect squares too? $\endgroup$ – Srivatsan Sep 21 '11 at 18:49
  • $\begingroup$ @JasonDeVito: Correction!! I was completely wrong, you had a very good point. The $l=2$ case is much, much easier and was solved 36 years early in 1939 by Erdos! $\endgroup$ – Eric Naslund Sep 21 '11 at 18:50
  • $\begingroup$ Thanks for the links, Martin and Eric. $\endgroup$ – francis-jamet Sep 21 '11 at 19:10

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