Find all real numbers $x$ and $y$ such that $x+y=3$ and $x^{5}+y^{5}=33$

Hint: Use the relations between coefficients and roots.
We need to start by expressing the quantity $x^{5}+y^{5}$ in terms of the elementary symmetric polynomials $(x+y)$ and $x y$. The most natural approach is to use the binomial expansion:

$$
\begin{aligned}
3^{5}=(x+y)^{5} & =x^{5}+5 x^{4} y+10 x^{3} y^{2}+10 x^{2} y^{3}+5 x y^{4}+y^{5} \\
& =33+5 x y\left(x^{3}+2 x^{2} y+2 x y^{2}+y^{3}\right) \\
& =33+5 x y\left((x+y)^{3}-x y(x+y)\right) \\
& =33+5 \cdot 3^{3}(x y)-5 \cdot 3(x y)^{2}
\end{aligned}
$$

We can now solve the following equation in $z=x y: 5 \cdot 3 z^{2}-5 \cdot 3^{3} z+3\left(3^{4}-11\right)=0$ or $z^{2}-9 z+14=0$, whose solutions are $z=7$ and $z=2$. We then have two cases:

- $x, y$ are the roots of $X^{2}-3 X+7$, which are not real since $3^{2}-4 \cdot 7<0$.
- $x, y$ are the roots of $X^{2}-3 X+2$, which are 1 and 2, so $(x, y)=(1,2)$ or $(x, y)=(2,1)$.

There are finally two solution pairs, which are $(1,2)$ and $(2,1)$.