Question

p is an odd prime. If p | x^p+y^p, then p^2 | x^p+y^p. Any helpful hints to prove this? I know it involves a contradiction with Fermat's Little Theorem.

Answer 1

Jumping into a meeting, but am eager to see if you have any insight #Zarkon.

Answer 2

I have an idea...it uses Fermat's little theorem, but it is a direct proof

Answer 3

So, assume \[p | x^p+y^p\] and show \[p^2 | x^p+y^p\], directly. Is it trivial?

Answer 4

it's not too bad i don't think

Answer 5

start by using fermat on \(x^p\) and \(y^p\)

Answer 6

oh

Answer 7

\[x^p\equiv x~ (mod~ p)\]
\[y^p\equiv y~ (mod~ p)\]

Answer 8

\[(x^p+y^p)\equiv x+y~ (mod~ p)\]

Answer 9

right?

Answer 10

yes

Answer 11

raise to the power p?

Answer 12

\[p|x^p+y^p\]

so

\[p|x+y\]

Answer 13

since \(x^p+y^p\) and \(x+y\) are congruent mod \(p\)

Answer 14

ok?

Answer 15

yes. thank you. I was over thinking the contradiction approach.

Answer 16

if \(p|x+y\) then \(x+y=kp\)

for some integer \(k\)


therefore

\(x=kp-y\)

Answer 17

then look at

\[x^p+y^p=(kp-y)^p+y^p\]

expand with the binomial theorem

Answer 18

then use the fact that p is odd to simplify

Answer 19

thank you