Let's determine the relatively prime numbers $a$ and $b$ if

$$
\frac{a^{2}-b^{2}}{a^{3}-b^{3}}=\frac{49}{1801}
$$
$$
\frac{a^{2}-b^{2}}{a^{3}-b^{3}}=\frac{(a-b)(a+b)}{(a-b)\left(a^{2}+a b+b^{2}\right)}=\frac{a+b}{a^{2}+a b+b^{2}}
$$

Now, if $a$ and $b$ are relatively prime, then

$$
n_{1}=a+b \quad \text { and } \quad n_{2}=a^{2}+a b+b^{2}=(a+b)^{2}-a b
$$

must also be relatively prime numbers.

Let $p$ be a prime number that is a common divisor of $n_{1}$ and $n_{2}$; then $p$ must also be a divisor of

$$
n_{1}^{2}-n_{2}=a b
$$

which means $p$ must be a divisor of either $a$ or $b$. If $p$ is a divisor of, say, $a$, then it is also a common divisor of

$$
a \text { and } n_{1}=a+b
$$

and thus a divisor of $b$ as well. Therefore, if $n_{1}$ and $n_{2}$ are not relatively prime, then $a$ and $b$ are not relatively prime either; this contradicts our assumption. Hence, $n_{1}$ and $n_{2}$ must be relatively prime, meaning the fraction

$$
\frac{a+b}{a^{2}+a b+b^{2}}
$$

is irreducible. However, $\frac{49}{1801}$ is also irreducible. This implies that

$$
a+b=49 \quad \text { and } \quad(a+b)^{2}-a b=1801, \quad \text { i.e., } \quad a b=49^{2}-1801=600
$$

Thus, $a$ and $b$ are the roots of the equation

$$
x^{2}-49 x+600=0 \quad \text { which are } 25 \text { and } 24
$$

The solution to the problem is: $a=25, b=24$ or $a=24, b=25$.

Péter Forgács (Ág. ev. g. VI. o., Bp.)