$x!=y^n$ for $x,y \neq 0,1$

Question

A straightforward problem (find all integers such that $m!+3=n^2$) led me into thinking about the integers for which: $$x!=y^2$$ is true.

I argued that other than the trivial case ($x!=1$) that this can't be true. While not rigorous it went something like this:

In order for a number to be a perfect square the powers of its prime factors must be even. Now consider the largest prime in $x!$ which I shall call $k$.

$k$ can only appear once in the prime factorisation of $x!$ (assuming that $2k<x$, an assumption I can't seem to prove). Therefore $x!$ cannot be a perfect square.

This argument seems to apply for any integer power $n$ as in that scenario $k$ would have to appear $n$ times in the prime factorisation of $x!$, but I've shown that $k$ only appears once. Therefore for integer $x,y\neq 0,1\; ,$ $x!\neq y^n$.

So my questions are:

1) Is my proof correct?

2) How do I prove my assumption if so?

3) Are there any other (better) proofs for the $x!\neq y^2$ or $x!\neq y^n$?

Answer 1

1) Yes, this is correct.

2) The assumption you can't seem to prove, namely that $2k<x$, is true, but it is not a trivial result (which is why it isn't easy to prove it). It's known as Bertrand's postulate. The full statement is:

For any positive integer $n$, there exists a prime number lying between $n$ and $2n$.

3) There might be a way to prove this without using Bertrand's postulate, but I imagine it would either be incredibly tortuous or would involve some other deep result. As shown in the linked answer below, you can use some of the ideas from a well-known proof of Bertrand's postulate to prove that no factorial is a square. This proof is a little shorter than proving Bertrand's postulate and then using that result.

Update: Your question (3) was asked here:

$n!$ is never a perfect square if $n\geq2$. Is there a proof of this that doesn't use Chebyshev's theorem? .

Answer 2

Yes, the proof is correct. In other words, consider the largest prime $p$ not exceeding $x$. And if $2p<x$ then by Bertrand's postulate there exist a prime $p_1>p$, a contradiction. so $x!$ cannot be a prefect square or perfect power also.