Get ready Part 24

Number Theory Level 3

For all prime numbers $p$ greater than $3$ and $k=\left \lfloor \dfrac{2p}{3}\right \rfloor$ , is the sum below always divisible by $p^2$ ?

$\sum_{n=1}^k \binom pn = \binom{p}{1}+\binom{p}{2}+\binom{p}{3}+\dots+\binom{p}{k}$

Notations:

$\lfloor \cdot \rfloor$ denotes the floor function .
$\binom MN = \frac {M!}{N!(M-N)!}$ denotes the binomial coefficient .

Yes No

1 solution

Patrick Corn
Feb 26, 2019

Here is a thread with the answer.

The first step is to note that for $1 \le n \le p-1,$ $\frac1{p} \binom{p}{n} \equiv \frac{(-1)^{p-n}}n \equiv \frac{(-1)^{n-1}}{n}$ mod $p$ for odd primes $p.$ Showing that the sum of these is congruent to $0$ mod $p$ is pretty fiddly, and is in the link.

Get ready Part 24

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