[ [Equations in integers] [Factorization $]$

Specify all such natural $n$ and integers $x$ and $y$, not equal to each other, for which the equation holds: $x+x^{2}+x^{4}+$ $\ldots+x^{2^{n}}=y+y^{2}+y^{4}+\ldots+y^{2^{n}}$

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For $n=1$, the equation takes the form $x+x^{2}=y+y^{2}$, or $(x-y)(x+y+1)=0$, from which we immediately obtain the solution.

For $n \geq 2$, the equation can be rewritten as follows: $(x-y)\left(1+(x+y)\left(1+x^{2}+y^{2}+\ldots\right)\right)=0$, where all terms in the last parenthesis are non-negative. Since $x \neq y$, both sides can be divided by $x-y$. We get: $(x+y)\left(1+x^{2}+y^{2}+\ldots\right)=-1$. The second parenthesis must have an absolute value of 1, from which $x=y=0$, which contradicts the condition.

## Answer

$n=1, x=k, y=-k-1$ for all integers $k$.