We are to find the natural number $a$ such that $3a + 1$ is a divisor of $a^2 + 5$. Set $b = 3a + 1$. Then we have $$a^2 + 5 = \frac{b^2 - \mathbf{OO}b + \mathbf{PQ}}{\mathbf{R}}.$$ On the other hand, since $b$ is a divisor of $a^2 + 5$, $a^2 + 5$ can be expressed as $$a^2 + 5 = bc$$ for some natural number $c$. From (1) and (2), we have $$b(\mathbf{S}c - b + \mathbf{IT}) = \mathbf{UV}.$$ This shows that $b$ must also be one of the divisors of UV. Of these, only $b = \mathbf{WX}$ is a number such that $a$ is a natural number. Hence, $a = \mathbf{YZ}$.
We are to find the natural number $a$ such that $3a + 1$ is a divisor of $a^2 + 5$.
Set $b = 3a + 1$. Then we have
$$a^2 + 5 = \frac{b^2 - \mathbf{OO}b + \mathbf{PQ}}{\mathbf{R}}.$$
On the other hand, since $b$ is a divisor of $a^2 + 5$, $a^2 + 5$ can be expressed as
$$a^2 + 5 = bc$$
for some natural number $c$. From (1) and (2), we have
$$b(\mathbf{S}c - b + \mathbf{IT}) = \mathbf{UV}.$$
This shows that $b$ must also be one of the divisors of UV. Of these, only $b = \mathbf{WX}$ is a number such that $a$ is a natural number. Hence, $a = \mathbf{YZ}$.