Sum of binomials

**james_bond** · February 28th 2010, 10:49 AM

Let $p>3$ be a prime. Show that $p|\sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}$ .

**NonCommAlg** · February 28th 2010, 10:49 PM

Originally Posted by james_bond

Let $p>3$ be a prime. Show that $p|\sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}$ .

it's also true for $p=2.$ anyway, this is just a simple application of Lucas' theorem. by this theorem we have:

$\sum_{i=1}^p \binom{ip}{p} \binom{(p-i+1)p}{p} \equiv \sum_{i=1}^p \binom{i}{1} \binom{p-i+1}{1}=\sum_{i=1}^p i(p-i+1) \equiv \sum_{i=1}^p i(i-1)$

$=\frac{p(p^2 - 1)}{3} \equiv 0 \mod p,$ because every prime which is $\neq 3$ is in the form $3k \pm 1$ and thus $3 \mid p^2-1. \ \Box$

Remark: you might say that i didn't apply Lucas' theorem correctly for $i=1$ and $i=p.$ but actually what i did

is perfectly fine because:

if $i=1,$ then by Lucas' theorem: $\binom{(p-i+1)p}{p}=\binom{p^2}{p} \equiv \binom{1}{0} \binom{0}{1} = 0 \equiv \binom{p}{1}=\binom{p-i+1}{1} \mod p.$

similarly if $i=p,$ then by Lucas' theorem: $\binom{ip}{p}=\binom{p^2}{p} \equiv \binom{1}{0} \binom{0}{1} = 0 \equiv \binom{p}{1}=\binom{i}{1} \mod p.$

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