i really need help proving or disproving i don't know how?

Question

2x^n + 2y^n = z^n has no non-zero solutions x, y, z in Z when n > 2.
Z is integer.

Answer 1

without loss of generality we can assume that both x and y are not even

this is so because we can take a common factor

one atleast one is odd.

say x is odd and y even
lhs is not divisible by 4 but rhs is divisible by 8. which does not hold

say both are odd

(2k+1)^n mod 8 = 1 or 5
so
x^n mod 8 = 1 or 5
y^n mod 8 = 1 or 5

2x^n mod 8 = 2
2y^n mod 8 = 2
add LHS mod 8 = 4
RHS mod 8 = 0 if RHS is even or 1 or 6

which cannot hold
so no solution

Answer 2

Keep it that way; this has a very detailed proof.