Proving $\sum_{k=0}^{p-1} {k \choose l}$ is divisible by p

Question

Let $p$ be a prime. How do we prove the following?

$\sum_{k=0}^{p-1} {k \choose l}$ is divisible by p for $0\leq l \leq p-2$

I have brute-forced verified this for the special case $p=5$, but I fail to see a trick that will let me prove this for arbitrary prime $p$.

Just to give some context, I am attempting to solve this Irreducibility issue. If I can prove the above statement, I should be able to apply Eisenstein's Criterion and be done.

This is not clear. What exactly are you asking for, say, $p=5$? — lulu, Mar 02 '18 at 16:23
Prove what exactly? If I take $p=5,l=3$ what is it you are trying to show? — lulu, Mar 02 '18 at 16:25
Please, answer the specific question. For $p=5,l=3$ what is it you are trying to prove? You claim to have solved the problem for $p=5$...what is it you solved? — lulu, Mar 02 '18 at 16:28
Actually, I have made a horrible mistake in my question. Let me edit it. — Pascal's Wager, Mar 02 '18 at 16:28

score 3 · Accepted Answer · answered Mar 02 '18 at 16:42

3

Hint: $$\sum_{k=0}^{p-1} \binom{k}l = \binom{p}{l+1}$$ holds for all positive integers $p$. This can be proven by induction on $p$ and by using Pascal's identity.

answered Mar 02 '18 at 16:42

Mike Earnest

75,930

Proving $\sum_{k=0}^{p-1} {k \choose l}$ is divisible by p

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