<s>[INST] solving a Diophantine system of two equations Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-
\begin{align}
a^2 + b^2 & = c^3 \\
(a + b)^2 & = c^4
\end{align}
Thank you for your help.
 [/INST] First observe that $(a+b)^2=c^4 \implies a+b = c^2 , c>0$ Then we can combine that with $a^2+b^2=c^3$ to get $(a-b)^2=2c^3-c^4$. As $a-b \in N, (2c^3-c^4)^{1/2}= c(2c-c^2)^{1/2} \in N$ Which implies $c = 1,2$ . Checking the two values, $c=0$ has no natural solutions and $c=1$ has one natural solution of $a=2,b=2,c=2$.
If we include zero in the set of natural numbers then for the cases $c=0,1$ we have $(0,1,1)$ and $(1,0,1)$ as solutions. For the case $c=0$ we also have a fourth solution of $(0,0,0)$ 
</s>