Show that x^5 + y^5 |= z^5 if x,y,z are integers which are not divisible by 5.

Suppose that $\displaystyle x,y,z$ are integers, none of which is divisible by $\displaystyle 5$.

Show that $\displaystyle x^5 + y^5 \neq z^5$. Hint: work modulo $\displaystyle 25$.

I can only think of these congruences by Fermat's Little Theorem.

$\displaystyle x^5 \equiv x (\mod 5)$

$\displaystyle y^5 \equiv y (\mod 5)$

$\displaystyle z^5 \equiv z (\mod 5)$.

How can I get congruences in modulo 25 and prove the statement?

Is it proved by assuming that $\displaystyle x^5 + y^5 = z^5$ and then show a contradiction like the proof for $\displaystyle n = 5$ in

https://en.wikipedia.org/wiki/Proof_of_Fermat's_Last_Theorem_for_specific_expone nts ?

Re: Show that x^5 + y^5 |= z^5 if x,y,z are integers which are not divisible by 5.

Since:

(x+5)^5 mod 25 = x^5 (expand it out to prove)

and 1^5=1, 2^5=7, 3^5=18 and 4^5 = 24 (mod 25)

it follows that x^5, y^5 and z^5 = 1,7,18 or 24. for any x,y,z not divisible by 5.

Now we cannot add two numbers from 1,7,18 and 24 to get 1,7,18 or 24 so the equality x^5+y^5=z^5 is false.

Re: Show that x^5 + y^5 |= z^5 if x,y,z are integers which are not divisible by 5.

Thank you very much. Just writing out the reason for the last step for myself. If

$\displaystyle x^5 \equiv a \pmod{25}$

$\displaystyle y^5 \equiv b \pmod{25}$

$\displaystyle z^5 \equiv c \pmod{25}$

then

$\displaystyle x^5 + y^5 \equiv a + b \pmod{25}$

$\displaystyle z^5 \equiv c \pmod{25}$,

which means $\displaystyle a + b = c$ if $\displaystyle x^5 + y^5 = z^5$.