Find all strictly positive integers $a$ and $\mathrm{b}$ such that $a+1$ divides $a^{3} b-1$ and $b-1$ divides $b^{3} a+1$
We want $a^{3} b-1 \equiv 0 \mod a+1$. Since $a+1 \equiv 0 \mod a+1$, then $a \equiv -1 \mod a+1$, so $a^{3} \equiv (-1)^{3} \mod a+1$, hence $a^{3} \equiv -1 \mod a+1$. Substituting into $a^{3} b-1 \equiv 0 \mod a+1$, we get $-b-1 \equiv 0 \mod a+1$, which is $b+1 \equiv 0 \mod a+1$, so $a+1$ divides $b+1$.

On the other hand, we want $b^{3} a+1 \equiv 0 \mod b-1$. Since $b \equiv 1 \mod b-1$, then $b^{3} \equiv 1 \mod b-1$, so substituting into $b^{3} a+1 \equiv 0 \mod b-1$, we get $a+1 \equiv 0 \mod b-1$, which is $b-1$ divides $a+1$.

We have $b-1$ divides $a+1$ and $a+1$ divides $b+1$, so $b-1$ divides $b+1$, hence $b-1$ divides 2, which gives:

- 1st case: $b-1=2$, so $b=3$, which gives $a=3$ or $a=1$.
- 2nd case: $b-1=1$, so $b=2$, which gives $a=2$.

Finally, the solutions are $b=3$ and $a=1, a=b=2$ and $a=b=3$.

## 2 Monday, February 24th morning: Louise Gassot

## - Euclidean Algorithm and Bézout's Theorem -

Proposition 141 (Euclidean Algorithm). Let $a \geq b>0$ be two integers. We define a sequence $\left(u_{n}\right)$ by: $u_{1}=a, u_{2}=b$ and as long as $u_{n}$ is not zero, $u_{n+1}$ is the remainder of the Euclidean division of $u_{n-1}$ by $u_{n}$. Then the sequence $\left(u_{n}\right)$ necessarily stops and the last non-zero term of this sequence is $\operatorname{gcd}(a, b)$.