Why $xy=N\Rightarrow (((x−1)p−1\text{ mod }N)−1)\text{ mod} X=0$

Question

Experimentally, I found that for $N=xy;\ 3<x<y$ where $x,y$ are prime numbers and for prime numbers, $p>x$ the below expression is always true:

$$\Big(\big((x-1)^{p-1}\text{ mod }N\big) -1\Big) \text{ mod } {X} = 0$$

Why is it this way? And is it indeed true for all the $p,x$?

Answer 1

Since $p$ is an odd prime, we have $$(X-1)^{p-1}\equiv (-1)^{p-1}=1\mod X$$

Set $$u:=(X-1)^{p-1}\ modulo \ Y$$.

Suppose $$Xs\equiv 1\mod Y$$ and $$Yt\equiv 1\mod X$$ then $$q:=(Xsu+Yt)\ modulo \ N$$ satisfies $$q\equiv Yt\equiv 1\mod X$$ and $$q\equiv Xsu\equiv u\mod Y$$

The chinese remainder theorem states then $$(X-1)^{p-1}\equiv q\mod N$$ so $(X-1)^{p-1}$ modulo $N$ is $q$. Now $q-1\equiv q-YT\equiv 0\mod X$ , hence $X$ divides $q-1$ completing the proof.