Find the functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that, for all real numbers $x$ and $y$, we have:

$$
f\left(x^{2}+x+f(y)\right)=y+f(x)+f(x)^{2}
$$
First, it is clear that the function $\mathrm{f}: x \mapsto x$ is a solution. We will show that it is the only one.
In the following, we will denote by $E_{x, y}$ the equation $f\left(x^{2}+x+f(y)\right)=y+f(x)+f(x)^{2}$. First, $E_{0, y}$ indicates that $f(f(y))=y+f(0)+f(0)^{2}$, which shows that $f$ is both injective and surjective, hence bijective.
On the other hand, for all real numbers $x$ and $t$, note that $x^{2}+x=t^{2}+t$ if and only if $0=$ $x^{2}-t^{2}+x-t=(x-t)(x+t+1)$, that is, if $x=t$ or $t=-1-x$. In particular, by setting $t=-1-x$, then $E_{x, y}$ and $E_{t, y}$ indicate that $f(x)+(x)^{2}=f(t)+(t)^{2}$. Consequently, if $x \neq-1 / 2$, and since we then have $t \neq x$ and $f$ is injective, we also have $f(t)=-1-f(x)$ and $f(x) \neq-1 / 2$. We deduce in particular that $f(-1 / 2)=-1 / 2$.
We have just shown that $-1 / 2$ is a fixed point of $f$. Now, if $y$ and $x$ are fixed points of $f$, the equation $\mathbf{E}_{x, y}$ shows that $x^{2}+x+y$ is also a fixed point. Similarly, if $x$ and $x^{2}+x+y$ are fixed points of $f$, and since $f$ is bijective, then $y$ is also a fixed point. Here, by choosing $x=-1 / 2$, we observe that as soon as $y$ is a fixed point of $f, y-1 / 4$ and $y+1 / 4$ are also fixed points. In particular, 0 and $1 / 4$ are fixed points.
Under these conditions, $E_{x, 0}$ and $E_{0, y}$ show respectively that $f\left(x^{2}+x\right)=f(x)+f(x)^{2}$ and that $f(f(y))=y$. Then, if $t$ is a positive real number, let $x$ be a real number such that $x^{2}+x=t-1 / 4$. Then $\mathbf{E}_{x, 1 / 4}$ indicates that

$$
\mathbf{f}(\mathbf{t})=\mathbf{f}\left(\mathrm{x}^{2}+x+\mathbf{f}(1 / 4)\right)=1 / 4+\boldsymbol{f}(\mathrm{x})+\mathrm{f}(\mathrm{x})^{2} \geqslant 0
$$

Furthermore, if $z$ is a real number such that $z^{2}+z=\mathrm{t}$, then $\mathbf{E}_{z, \mathrm{f}(\mathrm{y})}$ indicates that

$$
f(t+y)=f(y)+f(z)+f(z)^{2}=f(y)+f\left(z^{2}+z\right)=f(y)+f(t)
$$

We deduce both that $f$ is increasing and that $f$ is in fact additive.
The function $f$ is therefore linear, that is, of the form $f: x \mapsto \lambda x$. We conclude by observing that, since $-1 / 2$ is a fixed point of $f$, it follows that $\lambda=1$, hence that $f$ is indeed the identity function.