1. Determine all pairs $(a, b)$ of natural numbers for which

$$
a+5 \sqrt{b}=b+5 \sqrt{a} .
$$
SOLUTION. By substituting $m=\sqrt{a}, n=\sqrt{b}$, we transform the equation into the form $m^{2}-n^{2}-5(m-$ $-n)=0$, from which, using the formula for the difference of squares, we get $(m-n)(m+n-5)=0$. Therefore, $m-n=0$ or $m+n=5$.

In the first case, after reverse substitution, we find that the problem is satisfied by all pairs of natural numbers $a, b$ for which $b=a$. In the second case, we get $\sqrt{a}+\sqrt{b}=5$. Thus, $1 \leqq \sqrt{a}, \sqrt{b} \leqq 4$, so it suffices to sequentially substitute $a=1,2, \ldots, 16$ into the relation

$$
b=(5-\sqrt{a})^{2}
$$

and check whether the corresponding number $b$ is natural.

The given equation does not change by swapping the unknowns $a, b$. We can therefore assume $a \leqq b$, which, together with the equality $\sqrt{a}+\sqrt{b}=5$, means that $\sqrt{a} \leqq 2.5$. Hence, $a \leqq 6.25$. Therefore, it suffices to limit the substitution to the values $a=1,2, \ldots, 6$ and determine the remaining solutions by swapping the numbers $a, b$ in the found pairs.

A more clever approach involves squaring the bracket on the right side of relation (1) and subsequently rearranging it to the form

$$
\frac{25+a-b}{10}=\sqrt{a}
$$

from which it is clear that the number $a$ (and, due to the symmetry of the given equation, also the number $b$) is a square of a natural number. (In the opposite case, the left side of equation (2) would be a rational number, while the right side would be an irrational number.) Then, the left side of relation (2) is a natural number less than five. Hence, the difference $a-b$ is an odd multiple of five. Assuming $a<b$, we then have either $(a, b)=(4,9)$ or $(a, b)=(1,16)$. The other two solutions arise by swapping the numbers $a, b$.

Conclusion: The given equation is satisfied only by the pairs $(a, b)=(1,16),(4,9),(9,4),(16,1)$ and all pairs $(a, a)$, where $a$ is any natural number.

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