How many positive factorials are also perfect Squares. So for example $1!=1=1^2$. How many others exist other than 1?

Is there any way to prove this?



Bertrand's postulate(in actuality a theorem) states that for every prime $p$, there exists another prime number between $p$ and $2p$. This means that $\forall n>1, n!$ will always have a single power of some prime number(s).

For any positive integer $n\ge 2$, there exists a prime $p$ such that $\frac{n}{2}<p\le n$ This implies that $p\mid n!$. For $n\ge 5, p^2>n$. So if $n!$ has unique prime factorization $p_1^{k_1} \cdot p_2^{k_2} \cdots p_r^{k_r} $, then the exponent of $p$ must be $1$. Since perfect squares must have even exponents in their prime factorizations, we know $n!$ cannot be a perfect square for $n \ge 5$.

  • $\begingroup$ @TheChaz2.0: No, there will be 2. $0!$ and $1!$. I'll leave it for you to see why. $\endgroup$ – Yagna Patel Nov 12 '15 at 16:19
  • $\begingroup$ While this is just a hint, I would recommend expanding your answer a bit, as the OP may not be very familiar with this. $\endgroup$ – Brevan Ellefsen Nov 12 '15 at 16:20
  • $\begingroup$ @BrevanEllefsen: How's that? $\endgroup$ – Yagna Patel Nov 12 '15 at 16:25
  • $\begingroup$ @yagnapatel Thank you! $\endgroup$ – Zachary Smith Nov 12 '15 at 16:51
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    $\begingroup$ And also, this question is a possible duplicate of: math.stackexchange.com/questions/12544/…. $\endgroup$ – Yagna Patel Nov 14 '15 at 1:01

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