## Task A-1.5.

Let $x$ and $y$ be distinct real numbers such that

$$
x+4=(y-2)^{2} \quad \text { and } \quad y+4=(x-2)^{2}
$$

Determine $x^{2}+y^{2}$.
## First solution.

Using the formula for the square of a binomial, the equations can be written as

$$
x=y^{2}-4 y \quad \text { and } \quad y=x^{2}-4 x
$$

If we subtract these two equations, we get

$$
x-y=y^{2}-x^{2}-4 y+4 x
$$

Applying the formula for the difference of squares, we get

$$
x-y=(y-x)(y+x)-4(y-x) \text {. }
$$

Since $x \neq y$, we can divide by $x-y$ and get $x+y=3$.

If we add the equations $x=y^{2}-4 y$ and $y=x^{2}-4 x$, we get

$$
x+y=y^{2}-4 y+x^{2}-4 x
$$

1 point

i.e., $x^{2}+y^{2}=5(x+y)$. 1 point

We conclude that $x^{2}+y^{2}=5 \cdot 3=15$. 1 point