MHB Prime numbers proof by contradiction

[tmt1]

For prime numbers, $a$, $b$, $c$, $a^2 + b^2 \ne c^2$. Prove this by contradiction.

So, I get that $a^2 = c^2 - b^2 = (c - b)(c +b)$

And I get that prime numbers are the product of 2 numbers that are either greater than one, or less than the prime numbers.

But I'm unsure how to go from here.

 

[kaliprasad]

For prime numbers, $a$, $b$, $c$, $a^2 + b^2 \ne c^2$. Prove this by contradiction.

So, I get that $a^2 = c^2 - b^2 = (c - b)(c +b)$

And I get that prime numbers are the product of 2 numbers that are either greater than one, or less than the prime numbers.

But I'm unsure how to go from here.

a is prime so either $c-b = 1$ and $c+b= a^2$ or $c-b=c+b=a$ can you proceed from here