1. For all real $x$, we have: $x^{2}-8 x+17=(x-4)^{2}+1$. Since $(x-4)^{2} \geq 0$, it follows that $(x-4)^{2}+1 \geq 1$, hence for all real $x$, $x^{2}-8 x+17>0$.

2. Suppose that $(x+2)^{2}-(x-3)^{2} \geq 0$. We write: $(x+2)^{2}-(x-3)^{2}=5(2 x-1)$ (third remarkable identity).

Thus, we have: $5(2 x-1) \geq 0$ so $2 x-1 \geq 0$ (since $5>0$). Therefore, $x \geq \frac{1}{2}$.

3. For $n=5, 6 n^{2}-7=6 \times 25-7=141=11 \times 13$. Therefore, we have shown that there exists at least one natural number $n$ such that 11 divides $6 n^{2}-7$.

## 2 Tuesday afternoon, 18th: Cécile Gachet

Introduction. This course provides exercises on the application of simple induction, and then presents some variants: double induction or more generally of order $k \in \mathbb{N}^{*}$, strong induction, and descending induction. Finally, the related principle of infinite descent is briefly explained.

Let's start with some exercises to check that everything is fine with simple induction: