Determine $2^{m}-1 \wedge 2^{n}-1$.

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Let's write the Euclidean division of $n$ by $m: n=q m+r$. Now we want to perform the Euclidean division of $2^{n}-1$ by $2^{m}-1$:

$$
2^{n}-1=\left(2^{m q}-1\right) 2^{r}+2^{r}-1=\left(2^{m}-1\right)\left(1+2^{m}+2^{2 m}+\cdots+2^{(q-1) m}\right)+2^{r}-1
$$

Moreover, $2^{r}-1<2^{m}-1$. Thus, the Euclidean algorithm associated with $2^{n}-1$ and $2^{m}-1$ can be performed in parallel with that of $n$ and $m$. If $d=n \wedge m$, we will have

$$
2^{m}-1 \wedge 2^{n}-1=2^{d}-1
$$

The theorem of Bézout: the last game we will play consists once again in retracing the Euclidean algorithm, in the manner of what we did at the end of the previous proof. We know that

$$
d=r_{l}=r_{l-2}-r_{l-1} q_{l-1}=r_{l-2} u_{1}+r_{l-1} v_{1}
$$

We then replace $r_{l-1}$ by its value given by the previous line:

$$
r_{l}=r_{l-2}-\left(r_{l-3}-r_{l-2} q_{l-2}\right) q_{l-1}=r_{l-3}\left(-q_{l-1}\right)+r_{l-2}\left(1+q_{l-2} q_{l-1}\right)=r_{l-3} u_{2}+r_{l-2} v_{2}
$$

And so on, at each step, we find two integers (relative) $\mathfrak{u}_{\mathrm{k}}$ and $v_{\mathrm{k}}$ such that $\mathrm{d}=$ $r_{l-k-1} u_{k}+r_{l-k} v_{k}$. Thus, at the end, we obtain two relative integers $u$ and $v$ such that $d=u r_{0}+v r_{1}$.

We have just shown half of the famous theorem of Bézout: two integers $a$ and $b$ are coprime if and only if there exist two relative integers $u$ and $v$ such that $1=a u+b v$.

Indeed, the fact that if $a \wedge b=1$ then there exist $u$ and $v$ comes from the analysis made below when $d=1$. The converse must now be easy. If there exist $u$ and $v$ such that $1=a u+b v$, and if $d$ divides $a$ and $b$, then it divides $a u+b v$ so $d=1$.