1

The specific question is find all solutions for

$(x+y)^{2012} = x^{2012} + y^{2012}$ but obviously, the number isn't "important."

It would be better to find a general solution for

$(x+y)^{n} = x^{n} + y^{n}$

It is immediately apparent that one possible solution is $(0,0)$ and taking this further

$(0,m)$ and $(m,0)$ where $m$ is an integer when $n$ is even.

However, what about the cases where both $(x,y)$ are not zero?

Possibly $(i,j) \neq 0$

Is this even possible?

VladeKR
  • 433
  • If you are looking for integer solutions, Fermat's Last Theorem will limit $n$. For positives, $(x+y)^n > x^n + y^n$ is another hurdle. Are you looking for real numbers? – Macavity Oct 20 '14 at 08:57
  • In short, yes.Well if we consider complex numbers, then through rigorous calculation/computation we will find the answers. But if the solution is strictly real, then how do we find it? – VladeKR Oct 20 '14 at 08:58
  • 1
    If both $(x,y)$ are not zero then divide through by $x^n$ and let $z = y/x$. Can you see why equality can't hold for $z>0$? Can you see why it's false for $-2\le z<0$? (assuming $n$ is even) Can you see why it's false for $z < -2$? – Erick Wong Oct 20 '14 at 15:19

0 Answers0