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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.

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    Isn't this reducible via completing the square to Pell's equation in the case where $k=1$?2011-09-21
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    @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 $n=0$.2011-09-21
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    (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.2011-09-21
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    I see. According to Wikipedia: Pell's equation is any Diophantine equation of the form $$x^2-ny^2=1$$ where $n$ is a _nonsquare_ integer. I missed the "nonsquare" part.2011-09-21
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    cant you just use bertrand postulate an the fact that the largest prime less than n+k divides the product?2011-09-22
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    ops my fact is wrong..sorry2011-09-22

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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.

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    Thanks for your answer, Eric.2011-09-21
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    Here's the link to Erdos 1975 paper: http://www.renyi.hu/~p_erdos/1975-46.pdf Some other papers from http://www.renyi.hu/~p_erdos/Erdos.html might be of interest too.2011-09-21
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    @Jason Can you explain your final comment? What convinced you that the problem is difficult for perfect squares too?2011-09-21
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    @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!2011-09-21
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    Thanks for the links, Martin and Eric.2011-09-21
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    Just a quick question regarding something in the beginning of the paper. Why must the $a$'s have only prime factors less than $k$? I couldn't follow the reasoning in the bracket.2011-09-21
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    @EuYu: All that he is doing is writing $n+i=a_i x_i^2$, that is splitting each factor into its square part (the $x_i$) and its squarefree part (the $a_i$). The next thing he says is that each $a_i$ must have all its factors less then $k$. The reason is the product could not be a square if there was some prime greater than $k$ dividing $a_i$, because such a prime cannot occur anywhere else in the product. (So we conclude primes larger than $k$ must appear to an even power)2011-09-21
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    @Eric Ah, thanks for the clarification.2011-09-21
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    @Srivatsan: After my initial question, Eric responded saying something to the effect of "most of the work goes into the square case", so I responded with a thank you and bowed out. I then saw that Eric deleted that comment, so I deleted mine. Apparently there was more to the story (and also apparently "@Jason" doesn't work once I no longer have comments in the threads). Sorry to cause confusion!2011-09-21