9
$\begingroup$

Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?

The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?

$\endgroup$
  • 3
    $\begingroup$ No (well, other than $n=1$). By Bertrand, there is a prime $p$ between $\frac n2$ and $n$ and that prime can only divide $n!$ to the first power. $\endgroup$ – lulu Sep 18 '16 at 0:51
11
$\begingroup$

This is impossible due to Bertrand's postulate, since there will always be a prime $ p $ in $ n! $ occuring with multiplicity $ 1 $ as long as $ n \geq 2 $. This actually implies that $ n! $ is never a perfect power for $ n \geq 2 $.

| cite | improve this answer | |
$\endgroup$
8
$\begingroup$

If $\sqrt[n]{n!} = k \in \mathbb{N}$ then $n! = k^n$. When $n\geq 2$ we have $2\mid n!$ so we must also have $2\mid k$ which means that we can write $k = 2^{m} \ell$ for some integers $m$ and $\ell$. This again means that

$$n! = 2^{mn}\ell^n \implies 2^{mn} \mid n!$$

so the power of two that divides $n!$ is $mn$ which is greater or equal to $n$. On the other hand the power of two that divides $n!$ can be computed as

$$\left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{4}\right\rfloor + \left\lfloor\frac{n}{8}\right\rfloor + \ldots$$

This expression is less than $\frac{n}{2} + \frac{n}{4} + \frac{n}{8} +\ldots = n$ which gives us a contradiction.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

If $n\gt1$ then $\sqrt[n]{n!}$ is not an integer (so it is an irrational number). A proof using Bertrand's postulate has been posted. The proof of Bertrand's postulate is somewhat involved. Here is a proof without using Bertrand's postulate.

For a prime number $p,$ the $p$-adic order of a natural number $m,$ denoted by $\nu_p(m),$ is the highest exponent $\nu$ such that $p^\nu$ divides $m;$ the number $m$ is a perfect $k^\text{th}$ power if and only if $\nu_p(m)$ is divisible by $k$ for every prime $p.$ We can show that $n!$ is not a perfect $n^\text{th}$ power (for $n\gt1$) by showing that $\nu_2(n!)$ is not divisible by $n;$ in fact, $0\lt\nu_2(n!)\lt n.$ The lower bound is obvious. For the upper bound, let $m=\lfloor\log_2(n)\rfloor$ and use Legendre's formula: $$\nu_2(n!)=\sum_{k=1}^\infty\left\lfloor\frac n{2^k}\right\rfloor=\sum_{k=1}^m\left\lfloor\frac n{2^k}\right\rfloor\le\sum_{k=1}^m\frac n{2^k}\lt\sum_{k=1}^\infty\frac n{2^k}=n.$$

A much more general (and difficult) result, the Erdős-Selfridge theorem, says that the product of two or more consecutive positive integers is never a perfect $k^\text{th}$ power for any $k\gt1.$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

In this answer, it is shown that the number of factors of $p$ that divide $n!$ is $$ \frac{n-\sigma_p(n)}{p-1}\tag{1} $$ where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.

For $n!$ be an $n^{\text{th}}$ power, $(1)$ must be a multiple of $n$ for any prime $p$.

For any $n\ge1$, we have $\sigma_p(n)\ge1$ for any prime $p$. Thus, $(1)$ is less than $n$, and since it must be a multiple of $n$, it must be $0$.

Thus, either $n=0$ or the number of factors of any prime $p$ that divides $n!$ must be $0$. Therefore, we either have $n=0$ or $n=1$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This is (for $n>2$) never an integer, since you always have a single prime number (meaning with exponent 1) and hence its $n$th root, namely the biggest prime smaller than $n$. Since the product of roots of different prime numbers are never integer, this number will never be an integer (for $n>2$).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

EDIT: The comment below by bof points out the flaw in my argument. I am not deleting this answer. This proves a much weaker statement version that instead of "never happening" this shows it cannot happen for two consecutive integers.


Bertrand's postulate is the shortest way to prove this as has been suggested in other answers/comments. Here is an alternative attempt.

Suppose $n!=x^n,$ and $(n+1)!=y^{n+1}$ for integers $x,y$. Then we get by dividing these two equations $$ n+1 =\bigg(\frac yx\bigg) ^n y $$ As LHS is an integer so is the RHS. The only way RHS can be an integer is $y/x$ be an integer (and at least 2). That means $n+1$ is bigger than $2^{n+1}$ (because y is at least 2). But this is absurd, giving the required contradiction.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You get a contradiction from assuming that $n!$ is an $n^\text{th}$ power and $(n+1)!$ is an $(n+1)^\text{st}$ power? How does that prove that $n!$ is never a perfect $n^\text{th}$ power? $\endgroup$ – bof Sep 18 '16 at 2:07
  • $\begingroup$ @bof: Your objection is valid. Let me think through this. $\endgroup$ – P Vanchinathan Sep 18 '16 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.