Proving a property of a prime number (Elementary number theory) Prove the following statement.
For all prime numbers $a$, $b$, and $c$, $a
^2 + b^2
\neq c^2$
.
What i tried
Proving by contradiction
Assume the negation of the statement
There exists prime numbers $a$, $b$, and $c$ such that $a
^2 + b^2
= c^2$
Since we know that all prime number except $2$ is also an odd number
 Then we let $a=2n+1$ $b=2l+1$ 
Then we have 
$4l^{2}+4l+1+4k^{2}+4k+1=c^{2}$
$4l^{2}+4l+4k^{2}+4k+2=c^{2}$
$2(2l^{2}+2l+2k^{2}+2k+1)=c^{2}$
From this expression,since $c^{2}$ is even then $c$ is even. But we have a case where $c=2$, so if $c=2$, then $c^{2}=4$ and 
$2l^{2}+2l+2k^{2}+2k+1=2$. Then 
$2l^{2}+2l+2k^{2}+2k=1$ and 
$l^{2}+l+k^{2}+k=0.5$
Which is not possible since the sum of integers have to also be an integer
Which contradicts with the face that $c$ is either odd or has a value of $2$ for 
it to be a prime number
Is my proof correct.Could anyone explain. Thanks

A more elementary approach: since the only quadratic residues $\pmod{3}$ are $0$ and $1$, if $a^2+b^2=c^2$ at least one number between $a$ and $b$ has to be a multiple of three. With a similar argument $\pmod{4}$ and $\pmod{5}$, we have that $a^2+b^2=c^2$ implies that at least one number between $a$ and $b$ is even, and at least one number among $a,b,c$ is a multiple of five. It follows that no triple of primes is a pythagorean triple.
