Find all polynomials $P \in \mathbb{R}[X]$ such that $16 P\left(X^{2}\right)=P(2 X)^{2}$

Hint: use the previous question
As $P(x)^{2}=16 P\left(\frac{x^{2}}{4}\right)$ is a polynomial in $x^{2}$, we can apply the previous question. In the first case, if there exists a polynomial $Q$ such that $P(x)=Q\left(x^{2}\right)$, we get $16 Q\left(x^{4}\right)=$ $16 P\left(x^{2}\right)=P(2 x)^{4}=Q\left(4 x^{2}\right)^{2}$, and thus $16 Q\left(x^{2}\right)=Q(4 x)^{2}$. In the second case, if there exists a polynomial $Q$ such that $P(x)^{2}=x Q\left(x^{2}\right)$, we similarly obtain that $4 Q\left(x^{2}\right)=Q(4 x)^{2}$. We can therefore reapply the previous question to $Q$, and similarly we get that for any integer $k \geqslant 0$, there exists an integer $0 \leqslant i \leqslant 2^{k}$ and a polynomial $R_{k}$ such that $P(x)=x^{i} R_{k}\left(x^{2^{k}}\right)$. By choosing $k$ such that $2^{k}>\operatorname{deg} P$, it follows that $R_{k}$ is necessarily constant and thus $P(x)=c x^{i}$. By plugging back into the initial equation, we get $P(x)=16\left(\frac{x}{4}\right)^{i}$ for some integer $i \geqslant 0$ (and all these solutions are valid, conversely).

## 4 Combinatorial Geometry (Rémi and Aurélien)

## Useful Ideas

1. Test small cases (as in all combinatorial problems)
2. Always try the most brute force method possible
3. Think about induction (strong
4. Order the points by increasing abscissa
5. Consider the convex hull
6. Triangulate polygons: to divide a polygon into triangles, you need to draw $n-3$ diagonals, resulting in $n-2$ triangles

## Exercises