Exercice 34

On note $u_{n} = 5^{2^{n}}$. Alors $u_{n + 1} - 1 = u_{n}^{2} - 1 = (u_{n} - 1)(u_{n} + 1)$. Comme $5 \equiv 1 \;[4]$, on a $u_{n} \equiv 1 \;[4]$, donc $u_{n} + 1 \equiv 2 \;[4]$ et $v_{2}(u_{n} + 1) = 1$. La proposition 10 (i) donne $v_{2}(u_{n + 1} - 1) = v_{2}(u_{n} - 1) + 1$, et $v_{2}(u_{0} - 1) = v_{2}(4) = 2$. $$\boxed{v_{2}\big(5^{2^{n}} - 1\big) = n + 2}$$ Vérification : $5^{2} - 1 = 24 = 2^{3} \times 3$ et $5^{4} - 1 = 624 = 2^{4} \times 39$. ✓