0

For a prime $p$, if $x^p \equiv y^p \pmod p $

Then prove that; $x^p \equiv y^p(\bmod p^2) $

Firstly I am confused where to approach to prove this question ,though i know several propositions and theorems related to congruences

Like by applying fermat's theorem we can easily derive $x\equiv y \pmod p $ But what to do after that ?

Secondly i approached with the proposition that ,iff $x \equiv y (\bmod n_i)$ And for $i=1,2$ Then $x \equiv y (\bmod [n_1,n_2]) $

But i am unable to prove the question with these two approaches

Please tell whether either of my approaches will help to prove the question if not then please give a proper proof with steps for the same

This is an olympiad book question(not a homework problem)

Akshaj Bansal
  • 1,028
  • 6
  • 10

2 Answers2

1

Hint: If $\ x\equiv y \left(\hspace{-0.7em}\mod p\right)\ $, then $\ x= y + np\ $ for some integer $\ n\ $. What happens if you you use the binomial theorem to expand $\ \left(y + np\right)^p\ $ in powers of $\ y\ $ and $\ np\ $?

lonza leggiera
  • 31,984
1

If $p=2$ then $x$ and $y$ have the same parity, hence the same quadratic residue $\pmod 4$. Otherwise p is odd, and

$$x^p-y^p=(x-y)\sum_{i=1}^{p}x^iy^{p-i-1}\equiv(x-y)\sum_{i=1}^{p}x^ix^{p-i-1}=(x-y)\sum_{i=1}^{p}x^{p-1}\equiv(x-y)\cdot p\cdot1 \equiv 0\pmod {p^2}$$

abc...
  • 5,010