Find all real numbers $x, y$ satisfying $x^{5}+y^{5}=33$ and $x+y=3$.

## Solutions
We set $\sigma_{1}=x+y$ and $\sigma_{2}=x y$. We observe that

$$
\sigma_{1}^{5}=x^{5}+y^{5}+x y\left(5 x^{3}+10 x^{2} y+10 x y^{2}+5 y^{3}\right)
$$

Furthermore,

$$
\sigma_{1}^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}=x^{3}+y^{3}+3 \sigma_{2} \sigma_{1}
$$

We deduce that

$$
\sigma_{1}^{5}=x^{5}+y^{5}+5 \sigma_{2}\left(x^{3}+y^{3}\right)+10 \sigma_{2}^{2} \sigma_{1}=x^{5}+y^{5}+5 \sigma_{2}\left(\sigma_{1}^{3}-3 \sigma_{1} \sigma_{2}\right)+10 \sigma_{2}^{2} \sigma_{1}
$$

Therefore,

$$
x^{5}+y^{5}=\sigma_{1}^{5}-5 \sigma_{2} \sigma_{1}^{3}+15 \sigma_{2}^{2} \sigma_{1}+5 \sigma_{2}^{2} \sigma_{1}
$$

Since $\sigma_{1}=3$, we deduce that

$$
33=15 \sigma_{2}^{2}-135 \sigma_{2}+243
$$

Thus, $\sigma_{2}^{2}-9 \sigma_{2}+14=0$, from which $\sigma_{2}=2$ or $\sigma_{2}=7$.

First case: $\sigma_{1}=3$ and $\sigma_{2}=7$. In this case, according to Vieta's formulas, $x$ and $y$ are solutions of

$$
z^{2}-3 z+7=0
$$

which has no real solutions (negative discriminant).

Second case: $\sigma_{1}=3$ and $\sigma_{2}=2$. In this case, according to Vieta's formulas, $x$ and $y$ are solutions of

$$
z^{2}-3 z+2=0
$$

and thus $\{x, y\}=\{1,2\}$.

The only solutions are therefore $x=1, y=2$ and $x=2, y=1$.

## 6 Friday afternoon, June 19: Thomas Budzinski

## Exercises on polynomials