Does the result given below work for composite numbers?

Question

So, in the example given below, author states for prime numbers. But in the proving part, we don't have any condition for p being prime. Now later on, this same result can be used for proving Fermat's little theorem. Now Fermat's theorem works for prime numbers (sometimes for composite also - Carmichael or non-charmichael). In this result why is author being specific for primes? It does work for composites also. So.is there any condition am I missing? Please check the image. Here

Answer 1

I posted a comment earlier but it seemed to have given you an incorrect idea of what I meant. Here is what I meant in detail:

Since, $$\binom{n}{r}\cdot r! = (n-r+1)\cdot (n-r+2)\cdots n$$

If $n$ is a prime, it readily follows that $n$ does not divide $r!$ and hence $n | \binom{n}{r}$.

Similarly you can argue if $n$ and $r!$ are coprime, then $n | \binom{n}{r}$, however this can be improved further.

If $n$ is a composite, you can write it as a product of primes each less than $n$ itself. Suppose $n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$.

Then for $n | \binom{n}{r}$, you will need to take into account $p$-adic valuations. Find the highest power of each prime factor of $n$ that divides $n!$, $(n-r)!$ and $r!$. I denote them by $v_{p_k}(n!)$, $v_{p_k}(n-r)!$, and $v_{p_k}(r!)$. If $v_{p_k}(n!)-v_{p_k}(r!) - v_{p_k}(n-r)! \geq a_k$, then ${p_k}^{a_k}$ divides $\binom{n}{r}$. You will have to find this for all $p$ which are factors of $n$.

As a special case, whenever $n$ and $r$ are co-prime, it must be true that $n | \binom{n}{r}$. You can find various proofs online. But this is not an exhaustive criterion.