(Croatia) Find all integer solutions of the equation 4x + y + 4√ xy − 28√ x − 14√ y + 48 = 0.

First, let \( \sqrt{x} = a \) and \( \sqrt{y} = b \), then \( x = a^2 \) and \( y = b^2 \). Substituting these into the original equation, we get:

4a^2 + b^2 + 4ab - 28a - 14b + 48 = 0

This can be rewritten as:

4a^2 + b^2 + 4ab - 28a - 14b + 48 = 0

We notice we need the equation to satisfy \( a \) and \( b \) being integers. Let us analyze when \( a \) and \( b \) take on whole number values.

Case: \( a=2 \)

4(2)^2 + b^2 + 4(2)b - 28(2) - 14b + 48 = 0 \implies 16 + b^2 + 8b - 56 - 14b + 48 = 0

We simplify this:

b^2 - 6b + 8 = 0

Solving this quadratic equation:

b = \frac{6 \pm \sqrt{36 - 4 \cdot 8}}{2} = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2}

Therefore, \( b \) can be:

b = 4 \quad \text{or} \quad b = 2

This leads us to solutions for:

\sqrt{x} = 2 \quad \Rightarrow \quad x = 4

And \( \sqrt{y} = 4 \Rightarrow y = 16 \) or \(\sqrt{y} = 2 \Rightarrow y = 4 \).

So we have:

(x, y) = (4, 16) \quad \text{and} \quad (4, 4)

Overall, integer solutions are:

(4, 16) \quad \text{and} \quad (4, 4)

The answer is:

(x, y) = (4, 16) \quad \text{and} \quad (4, 4)