Croatia_2018

Let $S=x+y$ and $P=xy$.

We are given:

$$S=x+y=x^2+y^2=x^3+y^3$$

Recall: $$x^2+y^2=(x+y{)}^2-2xy=S^2-2P$$ x^3 + y^3 = (x + y)^3 - 3 x y (x + y) = S^3 - 3 P S So, the system becomes: 1. $S = S^2 - 2P$ 2. $S = S^3 - 3 P S$ From (1): S = S^2 - 2P \implies S^2 - S - 2P = 0 \implies 2P = S^2 - S \implies P = \frac{S^2 - S}{2} $$From(2):$$ S = S^3 - 3 P S \implies S^3 - 3 P S - S = 0 \implies S^3 - S - 3 P S = 0 Substitute $P$ from above: S^3 - S - 3 S \cdot \frac{S^2 - S}{2} = 0 $$Multiplybothsidesby2:$$ 2 S^3 - 2 S - 3 S (S^2 - S) = 0 2 S^3 - 2 S - 3 S^3 + 3 S^2 = 0 (2 S^3 - 3 S^3) + (3 S^2) - 2 S = 0 (-S^3) + 3 S^2 - 2 S = 0 S^3 - 3 S^2 + 2 S = 0 S (S^2 - 3 S + 2) = 0 S (S - 1)(S - 2) = 0 So the possible values for $S$ are $0$, $1$, or $2$. Now, for each $S$, compute $P$: - If $S = 0$: P = \frac{0^2 - 0}{2} = 0 So $x$ and $y$ are roots of $t^2 - 0 t + 0 = 0 \implies t^2 = 0 \implies t = 0$ So $(x, y) = (0, 0)$ - If $S = 1$: P = \frac{1^2 - 1}{2} = 0 So $x$ and $y$ are roots of $t^2 - t = 0 \implies t (t - 1) = 0$ So $(x, y) = (0, 1)$ and $(1, 0)$ - If $S = 2$: P = \frac{2^2 - 2}{2} = \frac{4 - 2}{2} = 1 So $x$ and $y$ are roots of $t^2 - 2 t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1$ So $(x, y) = (1, 1)$ Therefore, all solutions are:(x, y) = (0, 0),\ (0, 1),\ (1, 0),\ (1, 1)$$