Proof of no primes such that $x^2 + y^2 = z^2$ I'm in a pretty simple "CS Math" course for year 1 Comp Sci, and I came across this:
Disprove, $x^2 + y^2 = z^2$, such that $x, y, z$ are primes
I thought of this as, if n is a prime, then prime factorization of n must be:
$n = z*1$
$n^2 = (z*1)(z*1) = (z)(z)$
So I tried to derive a contradiction:
Assume $x, y, z$ are primes such that $x^2 + y^2 = z^2$
Then: $x^2 = z^2 - y^2$
Then: $x^2 = (z-y)(z+y)$ #Contradiction, prime factorization of $x^2$ must be uniquely represented.
But then I noticed that this statement is true if $z-y = 1$ and $z+y = x^2$, So:
Then: $y = -1+z = y = z-1$
Because: $x^2 + y^2 = z^2 $
Then: $(z-1)(z-1) + z+y = z^2$
Then: $z^2 - 2z + 1 + z + y = z^2$
Then: $z^2 - z + y + 1 = z^2$ # subtract $z^2$ from both sides 
Then: $-z + y + 1 = 0$
Then: $y = -1 + z$ #Loop??
I remember reducing this to $x + y = z$, which I think is false?
EDIT: Thought of this:
z - y = 1
Then: z = y + 1
Then z must be 3, y must be 2 
Because: z is greater than y by 1, since z is a prime and y is a prime, z and y must be 3 and 2 respectively since no other prime number is 1 apart because one of them would be even if greater than 2, and any even number has at least 3 divisors (2, the even number, and 1) thus not a prime.
Then: $x^2 + 2^2 = 3^2$ ,Because $2^2 = 4, 3^2 = 9$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction: x is a prime, but sqrt(5) is not a prime!
Is this a solid evidence proof in that can I say that only prime numbers 1 apart are 2 and 3? I'm pretty sure there is a more simpler and intuitive way to prove this, are there any other ways to prove this via contradiction?

The easiest way is to note that they cannot all be odd, then show that the even one cannot be $2$.
Added:  Let $y=2$.  Then $z^2-x^2=(z+x)(z-x)=4$  As $z,x$ are both odd, $z-x \ge 2, z+x\ge 2$, which forces $z+x=2, z-x=2, z=2, x=0$ Contradiction.
