Croatia competition problems

Let $z = x + iy$, where $x, y \in \mathbb{R}$.

Then
$$|z - 5| = |(x + iy) - 5| = |(x - 5) + iy| = \sqrt{(x - 5)^2 + y^2}$$
and
$$|z - 1| = |(x + iy) - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2}.$$

The given equation is
$$\sqrt{(x - 5)^2 + y^2} = \sqrt{(x - 1)^2 + y^2} + 4.$$

Let $a = \sqrt{(x - 5)^2 + y^2}$ and $b = \sqrt{(x - 1)^2 + y^2}$, so $a = b + 4$.

Then
$$a^2 = b^2 + 8b + 16.$$
But $a^2 = (x - 5)^2 + y^2$ and $b^2 = (x - 1)^2 + y^2$.

So
$$(x - 5)^2 + y^2 = (x - 1)^2 + y^2 + 8b + 16.$$
The $y^2$ terms cancel:
$$(x - 5)^2 = (x - 1)^2 + 8b + 16.$$
Expand:
$$(x^2 - 10x + 25) = (x^2 - 2x + 1) + 8b + 16$$
$$-10x + 25 = -2x + 1 + 8b + 16$$
$$-10x + 25 + 2x - 1 - 16 = 8b$$
$$-8x + 8 = 8b$$
$$- x + 1 = b$$
But $b = \sqrt{(x - 1)^2 + y^2}$, so
$$- x + 1 = \sqrt{(x - 1)^2 + y^2}$$

Now, $\sqrt{(x - 1)^2 + y^2} \geq 0$, so $-x + 1 \geq 0$, i.e., $x \leq 1$.

Square both sides:
$$(-x + 1)^2 = (x - 1)^2 + y^2$$
But $(-x + 1)^2 = (x - 1)^2$, so
$$(x - 1)^2 = (x - 1)^2 + y^2$$
$$0 = y^2$$
$$y = 0$$

Therefore, $z = x + i \cdot 0 = x$ is a real number.

Thus, $z$ is real.