Find the polynomial functions $P: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $P\left(x^{2}+1\right)=P(x)^{2}+1$ and $P(0)=0$.


We construct a sequence $\left(a_{n}\right)$ recursively such that $a_{0}=0$ and $a_{n+1}=a_{n}^{2}+1$. Then, by induction, $P\left(a_{n}\right)=a_{n} \forall n \in \mathbb{N}$, hence $x \mapsto P(x)-x$ has an infinite number of roots. A polynomial function with an infinite number of roots must necessarily be the zero function, so the function we are looking for is $P: x \mapsto x$, which is easily verified to be valid.

Another similar method involves substituting the unknowns not with specific values but with expressions depending on one or more variables, often with the aim of allowing clever simplifications. Thus, it can be useful to choose $x$ in terms of $y$ (or vice versa), in a functional equation with two variables, to view it as an equation with a single variable, from a more restricted but sometimes more relevant perspective.

Thus, even though it is always good, as a first reflex, to test for example $f(0)$ and $f(1)$, some substitutions are more relevant than others: notably, $x \leftarrow x-f(y)$, $x \leftarrow f(y)$ and their countless variations can easily get rid of nested functions.

## Symmetry

Many functional equations are based on a principle of symmetry. Obviously, the notion of symmetry is very vague, and this method is therefore all the more adaptable to a wide variety of situations.

The first symmetry, the most classic, is that between $x$ and $y$ in each term. This particularly invites transformations of the form $(x, y) \leftarrow(u+v, u-v)$: decomposing $x$ and $y$ into a symmetric part $u$ and an antisymmetric part $v$, via a transformation that can easily be inverted by $(u, v) \leftarrow\left(\frac{x+y}{2}, \frac{x-y}{2}\right)$.

For three variables, we similarly have the Ravi transformation: $x, y$ and $z$ are then considered as the sides of a triangle, which we now express in terms of the distances between the vertices and the points of tangency with the inscribed circle: $(x, y, z) \leftarrow(a+b, b+c, c+a)$.

However, a situation even more interesting than the symmetry of an equation is the symmetry of one of its terms. Suppose, for instance, that the equation is written as $\xi(f, x, y)=$ $\psi(f, x, y)$, with $\psi$ symmetric with respect to $x$ and $y$ (i.e., $\psi(f, x, y)=\psi(f, y, x))$ but not $\xi$. Then we can deduce $\xi(f, x, y)=\xi(f, y, x)$ for any $f$ that is a solution, thus drastically reducing our search space.

More specifically, in situations of the form $\xi(f, x, y)=\xi(f, y, x)$, we can, for example, choose $(x, y) \leftarrow(x,-x)$ or $(x, y) \leftarrow\left(x, \frac{1}{x}\right)$ to preserve the symmetry, or conversely try to break it in some way. However, we should avoid setting $(x, y) \leftarrow(x, x)$, as the symmetry prevents obtaining any information from this substitution.