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.
 A: Hint: Since $(a^2+b^2)c=c^4=(a+b)^2$ we have that $a^2+b^2$ divides $(a+b)^2$. This implies $a=b$ for positive integers. Then we have $2a^2=c^3$, so that $(a,b,c)=(2,2,2)$.
A: 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 $(0,0,0)$ 
A: We have that 
$$
c=\frac{(a+b)^2}{a^2+b^2}=\frac{a^2+2ab+b^2}{a^2+b^2}=1+\frac{2ab}{a^2+b^2}, 
$$ 
so $\dfrac{2ab}{a^2+b^2}$  must be an integer. But by Cauchy inequality $2ab \leq a^2+b^2$ and an integer can be only if $2ab=a^2+b^2$. It implies that  that $a=b$ and $c=2$.
From the first equation we have $2 a^2=c^3=8 \implies a=2.$
Se we get the one solution $(2,2,2).$
If $0 \in \mathbb{N}$ then we have trivial  solution $(0, 0,0).$ Also there are two more cases when $2ab \leq a^2+b^2$ is integer for $ a=0 \implies b=1$ and for $b=0 \implies a=1.$ So we get also two solutions $(0,1,1)$ and $(1,0,1).$
