need arithmetic problem proof if  $a^2+b^2=a^3+b^3=a^4+b^4$
find a and b

If $a=0$, it's easy to see that $b=0,1$. If $b=0$, by the same way, we have $a=0,1$.
Now assume $ab\neq0$, let $b=xa$ for some $x\neq0$, then we have
\begin{eqnarray*}
x^2(a^2+1)=x^3(a^3+1)=x^4(a^4+1).
\end{eqnarray*}
So we get $(a^2+1)(a^4+1)=(a^3+1)^2$, that is, $a^2(a-1)^2=0$, which implies that $a=0$ or $a=1$. By the assumption, then $a=1$. Then we get $2x^2=2x^3=2x^4$, which implies that $x=1$, that is, $b=1$.
In summary, we have solutions $(a,b)=(0,0),(0,1),(1,0),(1,1)$.
