. Let $x, y \in \mathbb{R}$ be such that $x=y(3-y)^{2}$ and $y=x(3-x)^{2}$. Find all possible values of $x+y$.
. The set $\{0,3,4,5,8\}$ contains all possible values for $x+y$.

A pair $(x, x) \in \mathbb{R}^{2}$ satisfies the equations if and only if $x=x(3-x)^{2}$, and it is easy to see that this cubic equation has the solution set $\{0,2,4\}$. These pairs give us 0,4 and 8 as possible values for $x+y$.

Assuming $x \neq y$ let $s$ be the sum $x+y$ and $p$ be the product $x y$. Subtracting the first equation from the second and cancelling out the term $x-y$ we get

$$
p=s^{2}-6 s+10 \text {. }
$$

Adding the two equations gives

$$
0=s\left(s^{2}-3 p\right)-6\left(s^{2}-2 p\right)+8 s .
$$

Together the equations give

$$
0=s^{3}-12 s^{2}+47 s-60=(s-3)(s-4)(s-5) \text {. }
$$

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Solutions

The only possible values for $x+y$ when $x \neq y$ are therefore 3,4 and 5 . We have already seen that $x+y=4$ has a solution $x=y=2$.

Next we investigate the case $x+y=3$. Here we can simplify the given equations as $x=y x^{2}$ and $y=x y^{2}$. The number $x$ cannot be zero in this case, since otherwise $y$ and $k$ would also be zero. We can conclude that $x y=1$. The equations $x+y=3$ and $x y=1$, according to Vieta's Theorem, imply that $x$ and $y$ are the solutions of the equation $\lambda^{2}-3 \lambda+1=0$. Hence

$$
(x, y)=\left(\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right) \quad \text { or } \quad(x, y)=\left(\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right)
$$

and it is easy to verify that both pairs satisfy the equations. These pairs give us 3 as a possible value for $x+y$.

A simple calculation shows that if a pair $(x, y) \in \mathbb{R}^{2}$ satisfy the equations, then the pair (4-x,4-y) is solution to the equations. From the pairs we have just found, we can therefore construct pairs of solutions

$$
(x, y)=\left(\frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right) \text { and }(x, y)=\left(\frac{5+\sqrt{5}}{2}, \frac{5-\sqrt{5}}{2}\right),
$$

which give us 5 as a possible value for $x+y$.