Prove that for all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$ Question:
Prove that for all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$
My attempt:
Proof by contradiction:
Assume $ \ \exists a,b,c $ prime numbers such that $ \ a^{2} + b^{2} = c^{2}$. 
Then $ \ a^2 = c^2 - b^2 \implies a^2 = (c-b)(c+b) $.
I am not sure how to show a contradiction here. Could someone please tell me the easiest way to show a contradiction from here?
 A: To continue along the approach you've started: note that by unique factorization, you either have $c-b=1$, $c+b=a^2$ or $c-b=c+b=a$; both of these are impossible for prime $b$ and $c$ — can you see why?.
Alternately, you can use a sort of parity argument: exactly one of $a$, $b$, and $c$ must be $2$ (can you see why?). It trivially can't be $c$, so suppose it's $a$; then we have $4+b^2=c^2$. Now, how close can two squares be to each other?
A: Clearly the numbers $a,b,c$ can't all be odd. So let say $b$ is even and thus $2$ and $a=2x+1$ and $c=2y+1$. We get
 $$ x^2+x+1 = y^2+y$$ which give's us contradiction since left side is odd and right even. 
A: For any integer $a$, $a\equiv 0,1,2(\mod 3)\Rightarrow a^2\equiv 0,1,4(\mod 3)\Rightarrow a^2\equiv 0,1(\mod 3)$
Now $a,b,c$ are prime numbers and $a^2+b^2=c^2$. First, important thing to note that $a,b,c$ are distinct. 
If $a=3$, then $(c+b)(c-b)=9$. Check that $b,c$ can not prime numbers.
If $b=3$, then $(c+a)(c-a)=9$. Check that $a,c$ can not prime numbers.
If $c=3$, then $a^2+b^2\equiv 2(\mod 3)\Rightarrow c^2\equiv 2(\mod 3)$, which is impossible.
If none of $a,b,c$ is $3$, $a^2+b^2\equiv 2(\mod 3)$ and $c^2\equiv 2(\mod 3)$, which is impossible.
The conclusion is yours to decide.$\space\space\space\space\space\space\space\space\blacksquare$
