# 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?

• What if $x=-y$, and $n$ is odd? – Workaholic Jan 28 '15 at 21:15
• Sorry, I forgot to mention that $n$ is a natural number. – Lior Jan 28 '15 at 21:19
• We need to deal with the cases when one of $x$ or $y$ is positive and the other negative. – André Nicolas Jan 28 '15 at 21:21
• just taking $n=3$, it has many more solutions, like $(-1,1)$. You cannot conclude from the binomial that $x=0$ or $y=0$ is your error – user45150 Jan 28 '15 at 21:23
• I understand what you mean but how can we find roots if $x < 0$ and $y>0$. @user45150 Now I covered that case as well. – Lior Jan 28 '15 at 21:24

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 $y=0$.

• $x=y=-1$ and $n=1$, then $-2$ is definitely not positive. If $n=3$ then you get terms of different signs as well. – Asaf Karagila Jan 28 '15 at 22:23
• @Asaf If $x=y=-1$ then $-x-y=2$ is positive... – Myself Jan 28 '15 at 23:07
• Uhm, no, $(-x)+(-y)=(-1)+(-1)=-2$. – Asaf Karagila Jan 28 '15 at 23:08
• @Asaf $x=-1\implies -x = -(-1) = 1$? – Myself Jan 28 '15 at 23:09
• @Asaf good thing you didn't make that sound derogatory at all... Just bear with me for 5 seconds: if $x^n+y^n=z^n$ then $(-1)^nz^n=(-1)^nx^n+(-1)^ny^n$ so $(-x)^n+(-y)^n=(-z)^n$. Now put $z=x+y$ and observe that $-(x+y)=-x-y$ and obtain $(-x)^n+(-y)^n=(-x-y)^n$. Finally denote $X=-x$ and $Y=-y$ and obtain $X^n+Y^n=(X+Y)^n$. – Myself Jan 28 '15 at 23:25

You can do this purely geometrically, without much algebra. Let's get rid of $n=1$ case since every $x,y$ is a solution then.

Note that since both sides are of degree $n$, if $(a,b)$ is a solution then so is $(\lambda a, \lambda b)$ for any $\lambda \in \mathbb{R}$.

We know that $(0,y), (x,0)$ are solutions, so now assume that $x,y\neq 0$. Then by what we noted above, WLOG assume $y=1$, then we get $x^n +1 = (x+1)^n$. Now, draw two graphs of $x^n+1$ and $(x+1)^n$ and you see that: (i) when $n$ is even, the two graphs meet only at $(0,1)$, so we don't get any newer solutions; (ii) when $n$ is odd and $n>1$, the two graphs meet at $(0,1)$ and $(-1, 0)$, meaning that $x=-1$ is an additional solution.

In conclusion, we have that $x=0$, $y=0$, and if $n$ is odd additionally $(-a, a)$ as solutions.