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How can we prove that the product of $5$ consecutive integers cannot be a perfect square?

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No.${}{}{}{}{}$ –  Will Jagy Feb 14 '13 at 23:54
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Did you mean 5 consecutive natural numbers? Otherwise there's $0$, $1$, $2$, $3$, $4$ and product $0=0^2$. –  zaarcis Feb 15 '13 at 0:01
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@WillJagy: what does this 'No' refer to? –  Berci Feb 15 '13 at 0:06
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What if $a=0$ ? –  zaarcis Feb 15 '13 at 0:09
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@Berci the original version of the question simply said 'prove', not 'how can we prove?'. –  Steven Stadnicki Feb 15 '13 at 0:11
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I see no need to retype the answer given here, which is the first result when putting the title of this question into Google.

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Much more is known --- the product of two or more consecutive positive integers is never a power (meaning, a square or higher power). The paper by Erdos and Selfridge is freely available at renyi.hu/~p_erdos/1975-46.pdf –  Gerry Myerson Feb 15 '13 at 2:14
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