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

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    $\begingroup$ Did you mean 5 consecutive natural numbers? Otherwise there's $0$, $1$, $2$, $3$, $4$ and product $0=0^2$. $\endgroup$ – zaarcis Feb 15 '13 at 0:01
  • $\begingroup$ $a$$(a+1)$$(a+2)$$(a+3)$$(a+4)$ $\endgroup$ – user62189 Feb 15 '13 at 0:07
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    $\begingroup$ What if $a=0$ ? $\endgroup$ – zaarcis Feb 15 '13 at 0:09
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    $\begingroup$ This paper generalizes your question. $\endgroup$ – JimmyK4542 Aug 14 '14 at 21:23
  • $\begingroup$ Note that your product is divisible by $5!=120$ and hence by $3600$ if it is to be a square. $\endgroup$ – Mark Bennet Aug 14 '14 at 21:25

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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    $\begingroup$ 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 $\endgroup$ – Gerry Myerson Feb 15 '13 at 2:14

This paper of Erdős proves that the product of two or more consecutive numbers is never a square. This question was also asked before on Math.SE. You can also relate to Math Forum or google "product of five consecutive numbers is never a square" and variants to get many interesting results.


Let $N=n(n+1)(n+2)(n+3)(n+4)$. Then it is easily seen that $N=2^a3^b5^{2k_1}7^{2k_2}\cdots$ where $a\geq 2,b\geq 1, k_1\geq 1$ and $k_2,k_3,\dots\geq 0$.

We consider four possible forms of $n+i,0\leq i\leq 4$:

If $2| a,b$ then let $a,b=2m_1,2m_2$ so that $n+i=(2^{m_1}3^{m_2}5^{k_1}\dots)^2$

If $2|a, 2\nmid b$ then let $a,b=2m_1,2m_2+1$ so that $n+i=3(2^{m_1}3^{m_2}5^{k_1}\dots)^2$

Continuing, we find $2\nmid a, 2|b\implies n+i=2(2^{m_1}3^{m_2}5^{k_1}\dots)^2$ and $2\nmid a,b\implies n+i=6(2^{m_1}3^{m_2}5^{k_1}\dots)^2$

By the pigeonhole principle, at least two forms will occur in $N$. Clearly, the two forms bust have $2|a,b$ which implies that the sequence contains two squares. But the only sequence which does this is ${1,2,3,4,5}$ which does not multiply to a product. Thus, no five consecutive integers can multiply to a perfect square.


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