$a^2 +b^2 = c^2 \implies a^2 = (c-b)(c+b)$. Because $a$ is a prime number, therefore $$\left\{ \begin{array}{ll} 1 &= c-b \\ a^2 &=c+b \end{array} \right.$$ Hence we have $b+1=c$ and $a^2=c+b=2b+1<3b$. This leads to $\frac{a}{b}<\frac{3}{a}$. Because $a\ge3$, we have $\frac{3}{a}\le1$ which implies $\frac{a}{b}<1$, i.e. $a < b$.