Pure math problem

(1a) Assume both x,y are odd
Case 1: z is even
Let x=2m+1, y=2n+1, z=2k, where m,n,k are integers
(2m+1)^2 + (2n+1)^2 = (2k)^2
4m^2 + 4m + 1 + 4n^2 + 4n + 1 = 4k^2
4(m^2 + n^2 - k^2 + m + n + 1) = 2
=> 4 is a factor of 2 => contradiction

Case 2: z is odd (let z = 2k+1)
(2m+1)^2 + (2n+1)^2 = (2k+1)^2
4m^2 + 4m + 1 + 4n^2 + 4n + 1 = 4k^2 + 4k + 1
4(m^2 + n^2 - k^2 + m + n - k + 1) = 3
=> 4 is a factor of 3 => contradiction

=> At least one of x,y is even

(1b)
(i) x,y cannot be both odd (part a)
(ii) x,y cannot be both even (otherwise not relatively prime)
=> one of x,y is odd and other is even

(2)
Suppose n is not prime, then n=p x q (where both not equals to 1)
(2^n) - 1 = [2 ^ (pq)] - 1 = (2^p)^q -1
Let x = (2^p) [only for simplicity here]
(2^n) -1 = (x^q)-1
=(x-1)[x^(q-1) + x^(q-2) + ... + x + 1]
=> (x-1) = (2^p -1) is a factor of (2^n)-1
=> (2^n)-1 is not prime

Therefore, "n is a prime" is a necessary condition for"f(n) is prime"