(Awesome math 2015)(M) Find all functions $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f\left(a^{2}, f(b, c)+1\right)=a^{2}(b c+1) \quad \forall a, b, c \in \mathbb{R}
$$
We show that $f$ is injective in the second coordinate when we fix a non-zero first coordinate. If $x \neq 0$ and $f(x, a)=f(x, b)$, then we write $f(1, f(x, b)+1)=f(1++, f(x, a)+1)$ and thus $x(b-a)=0$. In particular, if $b c=b^{\prime} c^{\prime}$, we write that $f(1, f(b, c)+1)=f\left(1, f\left(b^{\prime}, c^{\prime}\right)+1\right)$, from which $f(b, c)=f\left(b^{\prime}, c^{\prime}\right)=g(b c)$.

We write in terms of $g$:

$$
g\left(a^{2}(1+g(b))=a^{2}(b+1)\right.
$$

and with $a=1$ we have

$$
\left.a^{2}(b+1)=g 1+g\left(a^{2}(b+1)-1\right)\right)
$$

and thus $a^{2}(1+g(b))=1+g\left(a^{2}(b+1)-1\right)$ since $g$ is bijective and therefore with $b=0$ and $b=-2$ we have that $g$ is piecewise affine, we write the correct synthesis and we have $f(a, b)=a b$.