Congruent Modulo P

Question

I need help proving that if $x$ is congruent $y$ mod $p$ and $u$ is congruent $v$ mod $p-1$ then $x^u$ is congruent $y^v$ mod $p$. I have tried breaking stuff down and writing it as multiple of $p$ but I am not getting anywhere with it. For example, $x-y =mp$ for some $m$ and $u-v =n(p-1)$. Any help is appreciated

Answer 1

Do you know Fermats Little Theorem?

If $p$ is prime and if $p\not \mid k$ then $k^{p-1}\equiv 1\pmod p$ and if $p|k$ then $k\equiv 0 \pmod p$ so $k^{p-1}\equiv 0 \pmod p$ and in either case $k^p \equiv k\pmod p$.

So your problem (assuming $p$ is prime-- it's not true otherwise):

First if $x\equiv y \equiv 0 \pmod p$ this is trivial $x^u\equiv y^v\equiv 0\pmod p$. So let's assume $x\equiv y \not \equiv 0\pmod p$. And by FLT $x^{p-1}\equiv 1\pmod p$ and $y^{p-1}\equiv 1\pmod p$.

Try to finish on your own before reading further:

As $u\equiv v\pmod{p-1}$ then $u = v + k(p-1)$ for some $k$ and

Try to finish on your own before reading further:

$x^u=x^{v+k(p-1)}= (x^{p-1})^kx^v\equiv 1^k*x^v\equiv x^v \equiv y^v \pmod p$.