3. Determine all pairs $(a, b)$ of natural numbers such that their difference $a-b$ is the fifth power of some prime number and for which $a-4 \sqrt{b}=b+4 \sqrt{a}$.

The written part of the school round in category C takes place

on Thursday, January 25, 2007

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphic display. These details will be communicated to the students before the start of the competition.

## Solutions to the written part of the school round in category $C$
3. From the equality $a-4 \sqrt{b}=b+4 \sqrt{a}$, we obtain the equality $a-b=4(\sqrt{a}+\sqrt{b})$. Since $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$, we get, after dividing by the positive number $\sqrt{a}+\sqrt{b}$, the equality

$$
\sqrt{a}-\sqrt{b}=4
$$

or

$$
\sqrt{a}=\sqrt{b}+4
$$

By squaring this, we get $a=b+16+8 \sqrt{b}$. Since the number $r=8 \sqrt{b}$ must be an integer, $\sqrt{b}$ is a rational root of a natural number, so $b=n^{2}$ for some natural number $n$. From the equality (2), we have $a=(n+4)^{2}$ and $a-b=(n+4)^{2}-n^{2}=2^{3}(n+2)$. The number $a-b$ is thus a fifth power of a prime number only when $n+2=2^{2}$ or $n=2$.

The only suitable pair $(a, b)$ is the pair $(36,4)$.

Note. When we realize after deriving the relation (1) that the expression in the parentheses on the right side of the equality $a-b=4(\sqrt{a}+\sqrt{b})$ is a positive rational, and thus a natural number, we see that $a-b=2^{5}$ must hold. For the roots $\sqrt{a}, \sqrt{b}$, we then get the system of two equations

$$
\begin{aligned}
& \sqrt{a}+\sqrt{b}=8 \\
& \sqrt{a}-\sqrt{b}=4
\end{aligned}
$$

Adding these equations yields $\sqrt{a}=6$ and subtracting them yields $\sqrt{b}=2$.

For a complete solution, award 6 points. The observation that the square root of a natural number is either irrational or a natural number does not need to be proven (see also the first auxiliary problem $\mathrm{k} 1$. of the home round).