1. If $x+y=1$ and $x^{2}+y^{2}=2$, what is $x^{4}+y^{4}$?
First solution.

Since $x^{2}+y^{2}=2$, we can rewrite this equation as

$(x+y)^{2}-2 x y=2$,

and from this equation, by substituting $x+y=1$, we get $x y=-\frac{1}{2}$.

Further, $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=$

$=4-2 \cdot \frac{1}{4}=$

$=4-\frac{1}{2}=$