For all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$ 3 (solution inside)

Question

Hey guys, I'm trying to figure out how this solution is moving forward, so far so good, but this specific line (which I've highlighted in the attached picture) makes like no sense. Does anyone have an idea what they are trying to say?

Answer 1

Suppose $a=b$. Then $2a^2=c^2$, which is impossible. Thus, without loss of generality, you can assume $a<b$. It also holds that $a<c$ and $b<c$. Then $a^2=(c-b)(c+b)$. Since $b$ and $c$ must be odd, the only possibility is that $a=2$. Thus $$ (c-b)(c+b)=4 $$ However, $b\ge3$ and $c\ge3$, so $c+b\ge6$.