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?
 A: 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.
A: If $x,y,z\in\mathbb{N}_{\geq 1}$ and $x^2+y^2=z^2$, then at least one number between $x$ and $y$ is even and at least one number between $x$ and $y$ is a multiple of three, since the only possibilities for a square $\!\!\pmod{3}$ or $\!\!\pmod{4}$ are to be $0$ or $1$. So, in order to prove the statement it is sufficient to check that $2^2+3^2$ is not a square.
A: I think this proof is sufficient:
If x, y, z are primes, $x^2 + y^2 \ne z^2$
There exists x, y, z that are primes such that $x^2 + y^2 = z^2$, derive contradiction
Then: $x^2 = z^2 - y^2$ 
Then: $x^2 = (z-y)(z+y)$
Then: $z - y = 1$ and $z + y = x^2$, because by the Fundamental Theorem of Arithmetic, the prime factorization of x which is a prime is:
$x = (x)(1)$
Then: $x^2 = (x)(1)(x)(1) = (x)(x)$
Then: $x^2$ has only three factors, $x^2, 1, x$. Since $x - y \ne x + y$, then by elimination process, $x-y = 1$ and $z + y = x^2$
Then: $x - y = 1$
Then: $x = y + 1$
Then: $x = 3, y = 2$
Because: x and y are primes, x is one greater than y implies that one of x or y must be even as they both cannot be odd (odd number - 1 cannot equal odd). The only even prime number is 2, so by deduction, $y = 2, z = 3$
Also: $z + y = x^2$, as shown above 
Then: $x^2 + 2^2 = 3^2$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction! Prime Numbers must be natural numbers, $\sqrt5$ is not a whole number or natural number
