# need arithmetic problem proof [on hold]

if a2+b2=a3+b3=a4+b4a^2+b^2=a^3+b^3=a^4+b^4

find a and b

=================

=================

2

=================

If a=0a=0, it’s easy to see that b=0,1b=0,1. If b=0b=0, by the same way, we have a=0,1a=0,1.

Now assume ab≠0ab\neq0, let b=xab=xa for some x≠0x\neq0, then we have
x2(a2+1)=x3(a3+1)=x4(a4+1).\begin{eqnarray*}
x^2(a^2+1)=x^3(a^3+1)=x^4(a^4+1).
\end{eqnarray*}

So we get (a2+1)(a4+1)=(a3+1)2(a^2+1)(a^4+1)=(a^3+1)^2, that is, a2(a−1)2=0a^2(a-1)^2=0, which implies that a=0a=0 or a=1a=1. By the assumption, then a=1a=1. Then we get 2×2=2×3=2x42x^2=2x^3=2x^4, which implies that x=1x=1, that is, b=1b=1.

In summary, we have solutions (a,b)=(0,0),(0,1),(1,0),(1,1)(a,b)=(0,0),(0,1),(1,0),(1,1).

sorry for my question , but Why you choose the case in which aa divides bb ? Indeed, I am not doubting about the final answer, I am just asking in order to logicaly find the proof of the final answer. So why not taking other cases ?
– Nizar
2 days ago

1

@Nizar — Zhao was looking for real solutions, so b=xab=xa at that point o the argument is OK. [No such thing as divisibility in reals..] Because there are no other real number solutions, there aren’t integer solutions either.
– coffeemath
2 days ago

okay, clear now ! Thanks @coffeemath
– Nizar
2 days ago

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 a2+b2=a3+b3a^2+b^2=a^3+b^3 (in red), with an isolated point: (0,0)(0,0). Note the invariance by symmetry (x,y)→(y,x)(x,y) \rightarrow (y,x).
the quartic curve (q) with equation a4+b4=a3+b3a^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)(1,0),(1,1),(0,1), to which must be added the isolated point (0,0)(0,0).

Note: the solution by @Mingfeng Zhao amounts to intersect these curves with a straight line b=tab=ta pivoting through the origin.