# Thread: More on Congruence mod primes

1. ## More on Congruence mod primes

Hi ... I'm going through the Zhao and Sun proof of "Some Curious Congruences Modulo Primes", and I'm stuck in several places, hehehe. In particular, I was wondering how the following congruence was shown.
For $i = 1,2,...,p-1$

$\frac{(-1)^{i-1}}{p} \left(\begin{array}{cc}p\\i\end{array}\right) = \frac{(-1)^{i-1}}{i} \left(\begin{array}{cc}p-1\\i-1\end{array}\right) \equiv \frac{1}{i} \ (mod \ p)$

I get the equality part, no problem. But it's not so clear (for me) how the congruence can be shown. Any help would be great ... I'm learning quite a bit

2. The congruence is clearly true for $p=2.$ Suppose $p$ is odd. We have

$p-1\equiv-1\,(\bmod\,p)$

$p-2\equiv-2\,(\bmod\,p)$

$\vdots$

$p-(p-i)\equiv-(p-i)\,(\bmod\,p)$

Hence

$i\cdot(i+1)\cdot\cdots\cdot(p-1)\equiv(-1)^{p-i}(p-i)!\,(\bmod\,p)\equiv(-1)^{i-1}(p-i)!\,(\bmod\,p)$

as $p$ is odd. Therefore

$(p-1)!\equiv(-1)^{i-1}(i-1)!(p-i)!\,(\bmod\,p)$

$\implies\ \frac{(-1)^{i-1}(p-1)!}{(i-1)!([p-1]-[i-1])!}\equiv1\,(\bmod\,p)$

and you can easily complete the proof from here.

3. Ah, thank you ... I thought it was related to set congruences somehow, but I wasn't too sure how. I should have spotted that though, hehehe ... but still, I wouldn't claim the congruence right away, it's not that obvious to me