Question

Find all the natural numbers $n$ such that $n^2$ does not divide $(n-2)!$ .

Answer 1

Suppose $n=pqr$, where $p<q$ are primes and $r>1$.

Then, $p\le 2,q\ge 3$ and $r\ge 2$, not necessarily a prime.

Thus, we have
$n-2\ge n-p=pqr-p\ge 5p>p$
$n-2\ge n-q=q(pr-1)\ge 3q>q$
$n-2\ge n-pr=pr(q-1)\ge 2pr>pr$
$n-2\ge n-qr=qr(p-1)\ge qr$
Observe that $p,q,pr,qr$ are all distinct.

Hence, there product divides $(n-2)$ !.

Thus, $n^2=p^2{q}^2{r}^2$ devides $(n-2)!$

In this case, $n=pq$ where $p,q$ are distinct primes $n=p^k$ for some $p$.
Case 1. Suppose $n=pq$ for some primes $p,q$ where $2<p<q$.

Then, $p\ge 3$ and $q\ge 5$ .
In this case,
$n-2>n-p=p(q-1)\ge 4p$
$n-2>n-q=q(p-1)\ge 2q$
Thus, $p,q,2p,2q$ are all distinct numbers in the set {1, 2, 3, ...., n - 2}.

We see that $n^2=p^2{q}^2$ divides $(n-2)!$.

We conclude that $n=2q$ for some prime numbers $q\ge 3$.

Note that, $n-2=2q-2<2q$ in this case so that $n^2$ does not divide $(n-2)!$
Case 2. Suppose $n=k^p$ for some prime $p$.

We observe that $p,2p,3p,...,(p^{k-1}-1)p$ all lie in the set {1, 2, 3, ...,$n-2$ }.

If $p^{k-1}-1\ge 2k$, then there are at least $2k$ multiples of $p$ in the set {1, 2, 3, ..., $n-2$ }.

Hence, $n^2=p^{2k}$ divides $(n-2)!$.

Thus, $p^{k-1}-1<2k$.

If $k\ge 5$, then $p^{k-1}-1\ge 2^{k-1}-1\ge 2k$, which maybe proved by an easy induction.

Hence, $k\le 4$.

If $k=1$, we get $n=p$, a prime.

If $k=2$, then $p-1<4$ so that $p=2$ or 3; we get

$n=2^2=4$ or $n=3^2=9$.

For $k=3$, we have $p^2-1<6$ giving $p=2$; $n=2^3=8$ in this case.

Finally , $k=4$ gives $p^3-1<8$.

Again, $p=2$ and $n=2^4=16$.

However, $n^2=2^8$ divides $14!$ and hence it is not a solution.
Thus, $n=2,2p$ for some prime $p$ or $n=8,9$.

It is easy to verify that these satisfy the conditions of the problem.