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?


Lincoln Blackham
  • 1,141
  • 1
  • 8
  • 3
  • 9
    "Isn't close at all"? All but finitely many cases is almost the entire problem: if you can get effective bounds on $N$ you only need to check finitely many cases. But the PNT is much stronger than Bertrand's postulate. – Qiaochu Yuan Apr 09 '11 at 20:12
  • I'll take your suggestion and work on finding a bound tonight. However, the strategy of my prime number theorem proof was to prove what is essentially Chebyshev's theorem just for primes. I'm not sure if this would count as not using Chebyshev's theorem, since my (eventual?) proof could be extended to the all natural numbers if the $N$ for both of them is the same. I'll check, and maybe they'll be different, but hopefully there's another way to prove this that uses more elementary methods. – Lincoln Blackham Apr 09 '11 at 21:42
  • @Qiaochu: Using a bound on $\pi(x)$ that I found, I was able to get an upper bound of 91 for $N$. However, as I said in my previous comment, my solution still petty much uses Chebyshev's theorem. I still want to see if there's a solution that doesn't use it in any form. – Lincoln Blackham Apr 12 '11 at 12:40
  • * You would like a prime $p$ such that $p\equiv -1 \mod 4$ and $n!\equiv -1 \mod p$, but unfortunately the residue of $n!+1$ is not $-1$ modulo 4 and all simple variations do not seem to work, either. – Phira Apr 18 '11 at 20:17
  • 2
    *Another possible approach: Can you find an argument that, say, $\binom {2k}{k}$ is not a square? – Phira Apr 18 '11 at 20:17
  • It really feels like there should be a much easier proof of this! I wonder if that's my wrong intuition? – quanta Apr 19 '11 at 16:07
  • 1
    could you please clarify what you mean by 'doesn't use' Chebyshev's theorem? Having tried this problem independently without using betrand's postulate or chebyshev's theorem, I've proved that the statement doesn't always hold true if there isn't a prime between (1/2)n and n and must hold true if (i.e. only if) chebyshev's theorem is true. But would a proof that said that there must be a prime between (1/2)n and (3/4)n work? What I'm trying to say is, if a proof presented implies this theorem (which it has to), it could also be derived from it. Does that still make it valid as an alternative? – jg mr chapb Sep 05 '15 at 10:00
  • 5
    Seems to be very hard to do without Chebyshev. Notice that $6! = 5 \cdot(12)^2$, so we must know that the prime $5$ exists between $3$ and $6$. – zhoraster Sep 09 '15 at 04:45
  • As gebra pointed out, this is linked closely to Chebychev's. Thus, you would have to find a different path to prove something stronger than Chebychev's, perhaps even entirely inclusive. The theorem is equivalent to stating that there is at least one prime factor with an odd degree. Starting with 3!, can you show by strong induction that any further accumulation of prime factors must have at least one with odd degree? Another way to look at it is, for any N, how many factors of 2 are in the integers < N? How many of 3? of 5? How could *all* of those be even at once? – Prune Sep 15 '15 at 20:21
  • I think it can be proved by prime number theorem. – Adesh Tamrakar Sep 24 '15 at 09:33
  • Hey for n=2, n! Is also not a perfect sqaure. – Adesh Tamrakar Oct 06 '15 at 17:35
  • 1
    The answer to your question is "yes", because there is a proof of Chebyshev's theorem that does not use Chebyshev's theorem. – bof Oct 22 '15 at 09:07

1 Answers1


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 $2$.

  • 971
  • 6
  • 15
Eric Naslund
  • 69,703
  • 11
  • 166
  • 260