Combinatorial Number Theory!

Number Theory Level 4

$\large{4^s \sum_{k=0}^{n-s} { 2s+2k-1 \choose 2s-1} \equiv \alpha \pmod p}$

Let $p=2n+1$ be a prime, where $n$ is an integer, and let $s$ be any integer such that $1 \leq s \leq n$ . If the above modular equation satisfies, where $\alpha$ is a positive integer less than $p$ , find the value of $\alpha$ .

The answer is 1.

1 solution

Daniel Branscombe
Jul 31, 2015

I'll be honest, I took a bit of a cheat shortcut.

In order for this problem to have a single solution then for all values of (n,s) which produce an a>0, then that a must be constant.

So I decided to try the first one, namely n=1 which gives p=3 and selected s=1 this collapse the sum to a single term, namely Binomial[1,1]=1 and thus we have

4^1 = 1 mod 3

thus a=1

Combinatorial Number Theory!

The answer is 1.

1 solution

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