A number theory problem by Áron Bán-Szabó

If $k\geq 6$ and $k$ is composite, then $k=pq$ , where $1<p<q<k-1$ or $k=p^2$ , where $1<p<k-1$ .

A) $k=pq\quad 1<p<q<k$

$(k-1)!=1\times 2\times\dots \times p\times \dots\times q\times \dots\times (k-1)$

So now the statement is definitly true .

B) $k=p^2\quad 1<p<k\quad p\geq3$

It is easy to see that now $2p<p^2-1$ , and

$(k-1)!=1\times 2\times\dots\times p\times\dots\times 2p\times\dots\times(k-1)$

So the statement is always true .

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It's easy to see that if $p$ is a prime number, then $p\not\mid (p-1)!$ but since Wilson's theorem $p\mid (p-1)!+1$

Recall the fundamental theorem of arithmetic : every integer $> 1$ is either prime or composite, in which case it is the product of a unique combination of smaller primes.

Furthermore, if a number $A$ divides another number $B > A$ , then all factors of $A$ will also divide $B$ . For instance, $10$ divides $20$ , so $1$ , $2$ , and $5$ must also divide $20$ . Any prime $p$ will only divide a number $N > p$ if multiples of $p$ also divide $N$ . This is why $p$ never divides $(p - 1)!$ , as seen in the numbers marked in blue.

Now consider a composite number $x$ with the prime factorization

$\displaystyle p_1^{q_1}p_2^{q_2}p_3^{q_3} \cdots p_k^{q_k}$

for primes $p_1, p_2, p_3 \ldots p_k < x$ . Now, it is elementary that $x \mid x! \ \forall \ x \geq 0$ . But is it true that $x \mid (x - 1)! \ \forall \ x \geq 0$ as well? Yes! , because $(x - 1)!$ is the product of all positive integers $\leq x - 1$ , which must necessarily contain its prime factors.