$$ a^{3} + b^{3} = (a+b)^{2} \Longrightarrow (a+b)( a^{2} -ab+ b^{2} )= (a+b)^{2} \Longrightarrow (n,-n),n \in Z are solutions.
Suppose (a+b )\neq 0 \Longrightarrow a^{2} -ab+ b^{2} =a+b \Longrightarrow a^{2} -(b+1)a+ b^{2} -b=0 \Longrightarrow \Delta =-3 b^{2} +6b+1 \Longrightarrow \Delta \geq 0 \Longrightarrow \frac{3-2 \sqrt{3} }{3} \leq b \leq \frac{3+2 \sqrt{3} }{3} \Longrightarrow b \in [0,1,2] \Longrightarrow b=0 \Longrightarrow a=1 \Longrightarrow (1,0) solution \vee b=1 \Longrightarrow a=0 \vee a=2 \Longrightarrow (0,1);(2,1) solutions \vee b=2 \Longrightarrow a=1 \vee a=2 \Longrightarrow (1,2);(2,2) solutions \Longrightarrow (1,0);(0,1);(2,1);(1,2);(2,2) and (n,-n) \wedge n \in N$$
