# Product of $5$ consecutive integers cannot be perfect square

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

• 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
• $a$$(a+1)$$(a+2)$$(a+3)$$(a+4)$ – user62189 Feb 15 '13 at 0:07
• What if $a=0$ ? – zaarcis Feb 15 '13 at 0:09
• This paper generalizes your question. – JimmyK4542 Aug 14 '14 at 21:23
• Note that your product is divisible by $5!=120$ and hence by $3600$ if it is to be a square. – 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.

• 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

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.