# Find n $\in \mathbb{N}^*$so that the equality is true for all $x, y \in \mathbb{R}$

I recently came across this problem...and it surpassed my understanding of maths. It goes as follows:

Find the n nonzero natural number, such that it satisfies the expression:

$$[(x+y)^n-(x^n+y^n)]^2=n^2x^2y^2(x+y)^2(x^2+xy+y^2)^{n-3}$$

where $$x,y \in \mathbb{R}$$

I noticed we could do something with the $$x^2+xy+y^2$$ term, so I tried to turn that term into a cube:

$$(x^2+xy+y^2)^{n-3}=\frac{(x-y)^{n-3}(x^2+xy+y^2)^{n-3}}{(x-y)^{n-3}}=\frac{(x^3-y^3)^{n-3}}{(x-y)^{n-3}}$$

And this is the point where I can't go on with simplifying terms. I might try squaring the LHS, but it wouldn't lead me nowhere in my opinion.

• Setting $(x,y)$ to $(1,1)$ gives three values for $n$. Then it's a matter of testing each.
– lulu
Commented Jan 31 at 13:02
• Hmm...but wouldn't those 3 n's be the only solutions? If it is true for all x and y then no matter what we choose, we would only get those 3 n's, rigth? Commented Jan 31 at 13:05
• Not following. The general $n$ must work for $(1,1)$, of course. But the fact that $n$ works for one point doesn't mean it holds generally. Therefore, the solution (if it exists) has to be one of those three values, but presumably the other two values fail. Still, checking for a particular $n$ shouldn't be hard.
– lulu
Commented Jan 31 at 13:07
• Note: checking the three candidates at some other point, like $(2,1)$ or whatever, ought to eliminate the bad values (if there are any) quickly. Then you only have to work with one candidate.
– lulu
Commented Jan 31 at 13:09
• Now it's clear...thank you very much! Commented Jan 31 at 13:12