23
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 4
    $\begingroup$ @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
  • 4
    $\begingroup$ (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
  • 1
    $\begingroup$ 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
25
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your answer, Eric. $\endgroup$ – francis-jamet Sep 21 '11 at 18:33
  • 4
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.