1. $b=0,c=a$ 2. $a=\frac{b(b-1)}{2}, c=\frac{b(b+1)}{2}$

${{a}^{2}}+{{b}^{3}}={{c}^{2}}\Leftrightarrow {{a}^{2}}+{{b}^{3}}={{(a+b)}^{2}}\Leftrightarrow {{b}^{3}}-{{b}^{2}}-2ab=0\Leftrightarrow b({{b}^{2}}-b-2a)=0$ and the rest is easy.