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

Solution 1:

Here is a way to do it. We'll need De Polignac's formula which is the statement that the largest $k$ such that $p^k$ divides $n!$ is $$k=\sum_{i}\left\lfloor\frac{n}{p^i}\right\rfloor.$$ Additionally, we'll take advantage of the fact that the function $\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$ is only ever equal to $0$ or $1$.

Proof: Let's start with even numbers. Suppose that $(2n)!$ is a square. Then $\binom{2n}{n}=\frac{(2n)!}{n!n!}$ is a square as well, and we may write $$\binom{2n}{n}=\prod_{p\leq2n}p^{v_{p}}$$ where each $v_p$ is even. The critical observation is that for primes $p>\sqrt{2n}$ we have $v_{p}=\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$, which must equal either $0$ or $1$, and since $v_p$ is even, we conclude that $v_{p}=0$ for $p>\sqrt{2n}$. This will lead to a contradiction as $\binom{2n}{n}$ cannot be composed of such a small number of primes - this would give impossibly strong upper bounds on the size of the central binomial coefficient.

For $p\leq\sqrt{2n}$, $$v_{p}=\sum_{i}\left\lfloor\frac{2n}{p^{i}}\right\rfloor-2\left\lfloor\frac{n}{p^{i}}\right\rfloor\leq\log_{p}2n$$ and so $p^{v_p}=\exp(v_p\log p)\leq\exp(\log(2 n))= 2n$, which gives the upper bound $$\binom{2n}{n}=\prod_{p\leq\sqrt{2n}}p^{v_{p}}\leq\left(2n\right)^{\sqrt{2n}}.$$ Expanding $(1+1)^{2n}$ there will be $2n+1$ terms of which $\binom{2n}{n}$ is the largest. This implies that $\binom{2n}{n}>\frac{2^{2n}}{2n+1}$, and since $$\frac{2^{2n}}{2n+1}>(2n)^{\sqrt{2n}}=2^{\sqrt{2n}\log_2(2n)}$$ for all $n> 18$, we conclude that $(2n)!$ is never a square.

To prove it for odd numbers, consider the quantity $\frac{(2n+1)!}{n!n!}$. Observing that $\left\lfloor\frac{2n+1}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$ only takes the values $0$ and $1$ for odd $p>1$ we see that the above proof carries through identically with a slight modification at the prime $2$.