Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers. 
How to prove that $a^2 + b^2 \neq c^2$?

The sum of two odd numbers are even, so one of the numbers must be $2$.
If $a$ or $b$ are $2$ we have $a^2+4=c^2$ or $4=(c+a)(c-a)$ Since $c-a$ and $c+a$ have the same parity, this is impossible.
If $c=2$ we have $a^2+b^2=4$ but since $a$ and $b$ are positive, both must be $1$, but $1^2+1^2=2$.
