need arithmetic problem proof if  $a^2+b^2=a^3+b^3=a^4+b^4$
find a and b
 A: 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 $(a,b)=(0,0),(0,1),(1,0),(1,1)$.
A: This is not a proof but a visualizing aid that cannot fit in a simple "remark". The graphics below displays: 


*

*the cubic curve (c) with equation $a^2+b^2=a^3+b^3$ (in red), with an isolated point: $(0,0)$. Note the invariance by symmetry $(x,y) \rightarrow (y,x)$. 

*the quartic curve (q) with equation $a^4+b^4=a^3+b^3$ (in black), with the same isolated point. Note the invariance by different symmetries.
One "sees" the three common points $(1,0),(1,1),(0,1)$, to which must be added the isolated point $(0,0)$. 
Note: the solution by @Mingfeng Zhao amounts to intersect these curves with a straight line $b=ta$ pivoting through the origin.

