0

Studying thoroughly a physics matrix system I found (I'm pretty sure that I'm not committing errors) that for $P$ prime number and $n$ any integer $n<P$, we can decompose, $$\binom{P}{n}=P \ q$$ Where $q$ is an integer that depending on $P$ and $n$ will change values.

My intuition tell me that this is very related to Fermat's little theorem, but looking bibliography I can not find this concrete expression or something analogue anywhere. So I'm searching for someone that help me to find a place where this theorem or corollary is shown, or a way to prove it.

Euler
  • 159
  • 1
    If $P$ is prime and $0 < n < P$ then ${P \choose n}=\frac{P!}{n!, (P-n)!}$ is an integer where the numerator is divisible by $P$ and the denominator is not – Henry Jun 02 '22 at 22:23
  • Okay, so $P q= P \frac{(P-1)!}{n!(P-n)!}$ And how do you know that the left number, $q=\frac{(P-1)!}{n!(P-n)!}$ is an integer (in the case that I'm right and this should be an integer) – Euler Jun 02 '22 at 22:30
  • 2
    Imagine a factorization of the denominator into primes. All the primes are smaller than $P$, hence could not cancel the $P$ in factorization of the numerator. Since we know that the binomial coefficient is an integer, the $P$ survives after all cancellation of primes in the denominator. – Sammy Black Jun 02 '22 at 22:49
  • 1
    It is related to Fermat's little theorem: Euler used the property you found to prove Fermat's little theorem by induction. See Section 2.3.1 on the Wikipedia page for Fermat's little theorem: https://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem. – KCd Jun 02 '22 at 23:14

1 Answers1

1

If $P$ is prime with the bound conditions, we have that ${P}\choose {n}$$=P \cdot \frac{(P-1)!}{n!(P-n)!}=Pq$. Obviously $P \mid Pq$, but only if $q \in \mathbb{Z}$.

As $p \in \mathbb{Z}$ , $Pq \in \mathbb{Z}$ , $P \cdot \frac{(P-1)!}{n!(P-n)!}=P \cdot \frac{(P-1)!\dots (P-n+1)}{k!}$, and $P$ coprime to all the factors of $k!$, $P$ has no contribution to make $Pq \in \mathbb{Z}$ which implies that $q$ must be an integer.

Ian Barquette
  • 130