Baltic Way 2021 · Problem 5

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

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

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 $x+y$.

Let $f(x)=x(3-x)^{2}$. It is easy to check that if $x<0$ then $f(x)<x$. In particular $f(f(x))<f(x)<x$ in this case, so that the pair $(x, f(x))$ cannot be a solution. Similarly, $f(x)>x$ if $x>4$, so the pair $(x, f(x))$ cannot be a solution in this case either.

Suppose that $(x, y) \in \mathbb{R}^{2}$ is a solution. According to the previous remark $x \in[0,4]$, and similarly, $y \in[0,4]$. Hence we may write $x=2+2 r$ and $y=2+2 s$ with $r, s \in[-1,1]$. After substitution and simplification, the equation $x=y(3-y)^{2}$ transforms into the equation $r=4 s^{3}-3 s$. Recall the trigonometric identities for threefold angles. If $s=\cos (\alpha)$ for some $\alpha \in \mathbb{R}$, then $r=4 \cos ^{3}(\alpha)-$ $3 \cos (\alpha)=\cos (3 \alpha)$. In the same way $s=4 r^{3}-3 r=\cos (9 \alpha)$.

We can deduce that $9 \alpha=2 \pi m+\alpha$ or $9 \alpha=2 \pi l-\alpha$ for some integers $m$ and $l$. In the former case we have $8 \alpha=2 \pi m$, so that $m \in\{0,1,2,3,4\}$, and the corresponding possible pairs of solutions can be found in Figure 1. In the former case we have $10 \alpha=2 \pi l$, so that $l \in\{0,1,2,3,4,5\}$, where $l=0$ and $l=5$ result in angles that we have already considered in the first case. We consider the other options in Figure 2 taking into account the well-known identities $\cos (\pi / 5)=(1+\sqrt{5}) / 4$ and $\cos (3 \pi / 5)=(1-\sqrt{5}) / 4$.

$m$	$8 \alpha$	$\alpha$	$r$	$s$	$x$	$y$	$x+y$
0	0	0	1	1	4	4	8
1	$2 \pi$	$\pi / 4$	$\sqrt{2} / 2$	$-\sqrt{2} / 2$	$2+\sqrt{2}$	$2-\sqrt{2}$	4
2	$4 \pi$	$\pi / 2$	0	0	2	2	4
3	$6 \pi$	$3 \pi / 4$	$-\sqrt{2} / 2$	$\sqrt{2} / 2$	$2-\sqrt{2}$	$2+\sqrt{2}$	4
4	$8 \pi$	$\pi$	-1	-1	0	0	0

Figure 1: Pairs of solutions and their sums

$l$	$10 \alpha$	$\alpha$	$r$	$s$	$x$	$y$	$x+y$
1	$2 \pi$	$\pi / 5$	$(1+\sqrt{5}) / 4$	$(1-\sqrt{5}) / 4$	$(5+\sqrt{5}) / 2$	$(5-\sqrt{5}) / 2$	5
2	$4 \pi$	$2 \pi / 5$	$(-1+\sqrt{5}) / 4$	$(-1-\sqrt{5}) / 4$	$(3+\sqrt{5}) / 2$	$(3-\sqrt{5}) / 2$	3
3	$6 \pi$	$3 \pi / 5$	$(1-\sqrt{5}) / 4$	$(1+\sqrt{5}) / 4$	$(5-\sqrt{5}) / 2$	$(5+\sqrt{5}) / 2$	5
4	$8 \pi$	$4 \pi / 5$	$(-1-\sqrt{5}) / 4$	$(-1+\sqrt{5}) / 4$	$(3-\sqrt{5}) / 2$	$(3+\sqrt{5}) / 2$	3

Figure 2: Pairs of solutions and their sums