# n!n! is never a perfect square if n≥2n\geq2. Is there a proof of this that doesn’t use Chebyshev’s theorem?

If $n\geq2$, then $n!$ is not a perfect square. The proof of this follows easily from Chebyshev’s theorem, which states that for any positive integer $n$ there exists a prime strictly between $n$ and $2n-2$. A proof can be found here.

Two weeks and four days ago, one of my classmates told me that it’s possible to prove that $n!$ is never a perfect square for $n\geq2$ without using Chebyshev’s theorem. I’ve been trying since that day to prove it in that manner, but the closest I’ve gotten is, through the use of the prime number theorem, showing that there exists a natural number $N$ such that if $n\geq N$, $n!$ is not a perfect square. This isn’t very close at all. I’ve tried numerous strategies over the past weeks and am now trying to use the Sylow theorems on $S_n$ to somehow show that $|S_n|$ can’t be square (I haven’t made any progress).

Was my classmate messing with me, or is there really a way to prove this result without Chebyshev’s Theorem? If it is possible, can someone point me in the right direction for a proof?

Thanks!

Here is a way to do it. We’ll need De Polignac’s formula which is the statement that the largest $$kk$$ such that $$p^kp^k$$ divides $$n!n!$$ is $$k=\sum_{i}\left\lfloor\frac{n}{p^i}\right\rfloor.k=\sum_{i}\left\lfloor\frac{n}{p^i}\right\rfloor.$$ Additionally, we’ll take advantage of the fact that the function $$\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$$ is only ever equal to $$00$$ or $$11$$.

Proof: Let’s start with even numbers. Suppose that $$(2n)!(2n)!$$
is a square. Then $$\binom{2n}{n}=\frac{(2n)!}{n!n!}\binom{2n}{n}=\frac{(2n)!}{n!n!}$$
is a square as well, and we may write $$\binom{2n}{n}=\prod_{p\leq2n}p^{v_{p}}\binom{2n}{n}=\prod_{p\leq2n}p^{v_{p}}$$ where each $$v_pv_p$$ is even.
The critical observation is that for primes $$p>\sqrt{2n}p>\sqrt{2n}$$
we have $$v_{p}=\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloorv_{p}=\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$$, which must equal either $$00$$
or $$11$$, and since $$v_pv_p$$ is even, we conclude that $$v_{p}=0v_{p}=0$$
for $$p>\sqrt{2n}p>\sqrt{2n}$$. This will lead to a contradiction as $$\binom{2n}{n}\binom{2n}{n}$$ 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 $$p\leq\sqrt{2n}p\leq\sqrt{2n}$$, $$v_{p}=\sum_{i}\left\lfloor\frac{2n}{p^{i}}\right\rfloor-2\left\lfloor\frac{n}{p^{i}}\right\rfloor\leq\log_{p}2nv_{p}=\sum_{i}\left\lfloor\frac{2n}{p^{i}}\right\rfloor-2\left\lfloor\frac{n}{p^{i}}\right\rfloor\leq\log_{p}2n$$
and so $$p^{v_p}=\exp(v_p\log p)\leq\exp(\log(2 n))= 2np^{v_p}=\exp(v_p\log p)\leq\exp(\log(2 n))= 2n$$, which gives the upper bound $$\binom{2n}{n}=\prod_{p\leq\sqrt{2n}}p^{v_{p}}\leq\left(2n\right)^{\sqrt{2n}}.\binom{2n}{n}=\prod_{p\leq\sqrt{2n}}p^{v_{p}}\leq\left(2n\right)^{\sqrt{2n}}.$$
Expanding $$(1+1)^{2n}(1+1)^{2n}$$
there will be $$2n+12n+1$$
terms of which $$\binom{2n}{n}\binom{2n}{n}$$
is the largest. This implies that $$\binom{2n}{n}>\frac{2^{2n}}{2n+1}\binom{2n}{n}>\frac{2^{2n}}{2n+1}$$, and since $$\frac{2^{2n}}{2n+1}>(2n)^{\sqrt{2n}}=2^{\sqrt{2n}\log_2(2n)}\frac{2^{2n}}{2n+1}>(2n)^{\sqrt{2n}}=2^{\sqrt{2n}\log_2(2n)}$$
for all $$n> 18n> 18$$, we conclude that $$(2n)!(2n)!$$ is never a square.

To prove it for odd numbers, consider the quantity $$\frac{(2n+1)!}{n!n!}\frac{(2n+1)!}{n!n!}$$.
Observing that $$\left\lfloor\frac{2n+1}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor\left\lfloor\frac{2n+1}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$$
only takes the values $$00$$ and $$11$$ for odd $$p>1p>1$$ we see that the above proof carries through identically with a slight modification at the prime $$22$$.