<s>[INST] Real solutions of $x^n + y^n = (x+y)^n$ I have to find all real solutions of the following equation:
$x^n + y^n = (x+y)^n$
Clearly for $n = 1$, the equation holds for every $x,y$ real numbers.
If $n$ is greater or equal to $2$, we do binomial expansion on RHS and from  $x^n + y^n - (x+y)^n = 0 $ it follows that the roots are either $x=0$ or $y=0$.
Am I missing out on something or these are the only solutions?
 [/INST] First, if $x=0$ or $y=0$ then the equation is satisfied.
Second, if $x>0$ and $y>0$ then because by binomial expansion
$$ (x+y)^n = x^n + y^n + \text{strictly positive terms}$$
the equation can never be satisfied. On the other hand, if $x<0$ and $y<0$ then the equation is equivalent to $(-x)^n+(-y)^n=(-x-y)^n$ where all terms are positive so this has no solutions by the previous case.
So we are left with the case where, without loss of generality $x>0>y$, or if we use $-y$ instead of $y$ as a variable, the case
$$ x^n + (-y)^n = (x-y)^n, \text{ where } x,y>0.$$
If $n$ is even then all terms are positive and clearly $x^n + (-y)^n = x^n+y^n > (x-y)^n = |x-y|^n$, because either $x>|x-y|$ or $y>|x-y|$. If $n$ is odd then, depending on the sign of $x-y$ the equation becomes
$$ x^n + (y-x)^n = y^n \text{ or } y^n + (x-y)^n = x^n,$$
with all three terms positive. This is itself of the form $X^n+Y^n = (X+Y)^n$ so the only solutions are given by $y=x$ (which means $x=-y$ in the original equation), $x=0$ and $y=0$.
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