Real sequences which sum to 0, multiply to 1.

Solution 1:

Use that

$$(x_n+y_n)^2-4x_ny_n=(x_n-y_n)^2.$$

So $(x_n-y_n)^2$ has a limit. If we take limits then

$$-4=0-4=\lim_{n\to\infty}(x_n+y_n)^2-4\lim_{n\to \infty}x_ny_n=\lim_{n\to\infty}((x_n+y_n)^2-4x_ny_n)=\lim_{n\to\infty}(x_n-y_n)^2.$$

But this gives a contradiction, since $(x_n-y_n)^2\ge 0,\forall n\in \Bbb{N}.$

So, the sequences do not exist.

Solution 2:

Assume toward a contradiction that there are such sequences $(x_n)_n$ and $(y_n)_n$. Because $\lim_n x_n y_n$ is positive, $x_n$ and $y_n$ eventually have the same sign. Without loss of generality, we may assume that for infinitely many $n$ they are both positive. By passing to a subsequence we may assume that $x_n$ and $y_n$ are both positive for all $n$. Then $x_n$ and $y_n$ are both between $0$ and $x_n + y_n$, which approaches zero as $n \to \infty$, so $x_n$ and $y_n$ themselves approach zero as $n \to \infty$. Therefore $\lim_n x_ny_n = 0$, which is a contradiction.