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In recent days, while I was doing exercises on combinatorics, I thought if a number $m!$ could be a perfect square. I proved to demonstrate it through the prime factorization. My attempt: $$k=m!=p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}....\cdot p_n^{a_n}$$ with $a_1,a_2,..,a_n$ that have to be even. Now if there exists a prime number $p_j$ between $\frac {k}{2}<p_j<k$ this prime number in the prime factorization the exponent ($a_j$) relative to the prime number $p_j$ is equal to $1$ ($a_j=1$) because the number $k$ isn't a multiple of $p_j$; for example $12!$ isn't a perfect square because the prime numbers $7$ and $11$ $6<7,11<12$ aren't divisors of $12$.

My question

There exist a prime number $p_j$ between $k$ and $\frac {k}{2}$ and how can I prove if there exist or not exist a number $m!=n^2$?

user26486
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Domenico Vuono
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5 Answers5

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Let $m\ge 3$ (otherwise $(m,n)=(0,\pm 1),(1,\pm 1)$).

Assume for contr. $\ n^2=m!\ $ for some $n\in\Bbb Z$.

Then $\ \exists p\in\Bbb P\ $ with $\ m/2<p<m\ $ by Berdrand's postulate.

$\,\Rightarrow\, p\mid m!\,\Rightarrow\, p\mid n^2\,\Rightarrow\, p^2\mid n^2\,\Rightarrow p^2\mid m!$. But $m<2p$, contr.

user26486
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Here's a sledge-hammer to crack this peanut: A question on primes and equal products

It references this:

Erdos and Selfridge, [The product of consecutive integers is never a power][1], Illinois J Math 19 (1975) 292-301.

The paper is here:

Link

Glorfindel
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marty cohen
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Suppose m! = n^2, m,n> 1. We are guaranteed that a prime exists between m!/2 and m! by Bertrand's postulate.Clearly,that prime can appear only once,but it is a factor of m! and therefore a factor of n^2.Since prime,it divides n,and so divides n^2 twice,which is impossible,and we have a contradiction.So m = n = 1 is the only solution. Edwin Gray

Edwin Gray
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Subtracting $m^2$ from both sides of your equation we get $$n^2-m^2=m!-m^2 \tag{0} $$ $$ (n-m)(n+m)=m\left((m-1)!-m\right), $$ that is, assuming $m\ne0$, $$ \frac{n}{m}+1=\frac{(m-1)!-m}{n-m}.\tag{1}$$ We do know that $n$ is a multiple of $m$, therefore both sides of $(1)$ are integer. Now, $(1)$ is equivalent to $$\frac{n}{m}=\frac{(m-1)!-n}{n-m} $$ $$\frac{n}{m^2}=\frac{\frac{(m-1)!}{m}-\frac{n}{m}}{n-m} \tag{2}$$ as well as $$\frac{n}{m^2}+\frac{1}{m}=\frac{\frac{(m-1)!}{m}-1}{n-m} \tag{3}.$$ But while both the right-hand sides of $(2)$ and $(3)$ are integer, since $(0)$ being rewritable as $$m^2\left(\left(\frac{n}{m}\right)^2 -1\right) =m( (m-1)! - m)$$ yields that $m$ divides $(m-1)!-m$, and so $(m-1)!$, their left-hand sides can be simultaneously integer if and only if $m=1$. Therefore, the only couples of solutions are $(m,n)=(0,\pm1),(1,\pm1)$.

Vincenzo Oliva
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  • How do you know $m\mid n$? – user26486 May 05 '15 at 22:55
  • @user31415, $m$ cannot be a prime ($p!$ is never a square since it's divisible by $p$ but not $p^2$), and for non-primes $m^2\mid m!$, so that $m^2\mid n^2\implies m\mid n$. What I don't see is why the right-hand side of $(3)$ is an integer. – Barry Cipra May 05 '15 at 23:06
  • @BarryCipra Because $(m-1)!-m$ is divisible by $m$, and we previously showed that $ \frac {(m-1)!-m}{n-m}'$ is an integer. – Vincenzo Oliva May 06 '15 at 05:00
  • @BarryCipra I see what I could have overlooked, please let me know if what I'm saying is incorrect. – Vincenzo Oliva May 06 '15 at 05:36
  • $(m-1)!-m$ is divisible by $m$ why? If $m$ is a prime number – Domenico Vuono May 06 '15 at 08:09
  • @DomenicoVuono Because $(m-1)!$ is. See my edit. If $m$ is prime that's impossible indeed, but that simply means your equality is false for all prime $m$. – Vincenzo Oliva May 06 '15 at 09:14
  • Yes right... Sei in classe? – Domenico Vuono May 06 '15 at 09:17
  • @DomenicoVuono Yep lol – Vincenzo Oliva May 06 '15 at 09:33
  • @VincenzoOliva, I'm still puzzled, because it looks like the only fact about $m!$ you've used is that it's of the form $m^2a$. That is, $m!=m(m-1)!$ and $m\mid (m-1)!$. I don't see where you've used the fact that a factorial is the product of all integers up to what's being factorialized. – Barry Cipra May 06 '15 at 13:55
  • @BarryCipra I guess I don't, but why is that alone a problem? – Vincenzo Oliva May 06 '15 at 14:13
  • @VincenzoOliva, it's a problem because your proof seems to apply in a much more general setting: If $f$ and $g$ are functions such that for any composite number $m$, $f(m)=mg(m-1)$ and $m\mid g(m-1)$, then $f(m)$ cannot be a square (for composite $m$). This is clearly not true for all such functions (in particular, take $g(k)=k+1$). I agree that the numerator in the right-hand side of $(3)$ is an integer, but the division by $m$ in it could conceivably leave a quotient that's no longer divisible by $n-m$. – Barry Cipra May 06 '15 at 14:36
  • @BarryCipra I get what you mean in the first part, but I think the real problem is what you point out in the end, and which I had suspected. In other words, if there weren't that mistake, the proof would be correct for this precise statement, and not the general one, no?

    I waited for other comments from you before deleting the answer, as I could be missing something. Do you think it's fixable? You know, my purpose was to prove the statement without Chebyshev's theorem.

     – Vincenzo Oliva May 06 '15 at 14:51
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    @VincenzoOliva, I think we're on the same page here. The first part of my previous comment was just a way to formalize what the problem was. I don't see any quick fix, but that hardly means there isn't one. It does seem that there ought to be a proof that doesn't use Chebyshev's theorem. Either that or else, maybe, assuming $m!$ is never a square (for $m\gt1$) gives a quick proof of Chebyshev! – Barry Cipra May 06 '15 at 15:17
  • @BarryCipra Thanks. Very kind, as always. – Vincenzo Oliva May 06 '15 at 15:52
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gcd is an associative binary operation. gcd(m,m-1)=1. Thus gcd of terms involved in m! is 1. It follows that m! cannot be a perfect square.

Adelafif
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