Find the pairs $(a, b)$ of non-zero natural numbers, with $b \neq 1$, such that the numbers $\frac{a^{3} b-1}{a+1}$ and $\frac{b^{3} a+1}{b-1}$ are non-zero natural numbers.

## 2 Solution
As $a^{3} b-1=b\left(a^{3}+1\right)-(b+1)$ and $a+1 \mid a^{3}+1$, we have $a+1 \mid b+1$. Meanwhile, as $b^{3} a+1=a\left(b^{3}-1\right)+(a+1)$ and $b-1 \mid b^{3}-1$, we have $b-1 \mid a+1$.

Thus, $b-1 \mid b+1$ and therefore $b-1 \mid 2$.

$\triangleright$ If $b=2$, then $a+1 \mid b+1=3$ gives $a=2$. Therefore, $(a, b)=(2,2)$ is the only solution in this case.

$\triangleright$ If $b=3$, then $a+1 \mid b+1=4$ gives $a=1$ or $a=3$. Therefore, $(a, b)=(1,3)$ and $(3,3)$ are the only solutions in this case.

In conclusion, $(a, b)=(1,3),(2,2)$ and $(3,3)$ are the only solutions.

## 3 Group C

## 1 Statement

Write the different problems on separate sheets. On each sheet, write your name, your group, and the problem number.

Rulers, set squares, and compasses are allowed. Protractors are prohibited. Calculators are prohibited, as are all electronic devices.