Wait this is equivalent to 2002 ISL A4, see the shorlist, page 29 for this exact question (with variables switched).

This is a normal problem for Cauchy Equation. Let g(x)=sqrt(f(x)) . By some Transformation we can get g(x+y)=g(x)+g(y) and g(xy)=g(x)g(y)

Let $P(x,y,z)$ be the assertion $(f(x)+1)(f(y)+f(z))=f(xy+z)+f(xz-y)$. $f\equiv 0$ works, conversely assume $f\not\equiv 0$. $P(0,0,0)$ implies $f(0)=0$ and $P(0,x,1)$ implies $f(x)=f(-x)$. And $P(x,y,0)$ shows $f$ is multiplicative. So $f(x)\geqslant 0$ for all $x$, so $P(x,x,1)$ gives (noting $f(1)=1$), $f(x^2+1)=(f(x)+1)^2\geqslant 1$. And so $f(x_1)\geqslant 1$ for all $x_1\geqslant 1$. Take $x>0$ and $y>1$, then $f(xy)=f(x)f(y)\geqslant f(x)$, so $f$ is increasing. So the function $e^{f(\ln x)}$ is additive and increasing, whence $f(x)=x^c$. Easy to see $c=2$ works only.

Let $P(x,y,z):=(f(x)+1)(f(y)+f(z))=f(xy+z)+f(xz-y)$ $P(0,0,0)$ yields $2f(0)(f(0)+1)=2f(0)\Longrightarrow f(0)^2+f(0)=f(0)\Longrightarrow f(0)^2=0$ thus $f(0)=0$ Furthermore $P(x,0,1)$ yields $(f(x)+1)f(1)=f(1)+f(x)\Longrightarrow f(x)f(1)=f(x)$ Thus either $f\equiv0$ which is one solution to the functional equation or $f(1)=1$ when $f\not\equiv 0$ So from now on assume that $f\not\equiv0$ $P(x,0,x)$ yields $f(x)(f(x)+1)=f(x)+f(x^2)\Longleftrightarrow f(x)^2+f(x)=f(x)+f(x^2)\Longrightarrow f(x)^2=f(x^2)$ $P(0,x,0)$ yields $f(x)=f(0)+f(-x)\Longrightarrow f(x)=f(-x)$ thus $f$ is even Notice that from $P(x,y,0)$ we obtain $f(y)(f(x)+1)=f(xy)+f(-y)=f(xy)+f(y)\Longrightarrow f(x)f(y)=f(xy)$ therefore $f$ is multiplicative. (Cauchy function) Moreover $P(x,x,1)$ yields $(f(x)+1)^2=f(x^2+1)$ however notice that $f(x^2+1)=(f(x)+1)^2\ge1$ which implies that $f(x^2+1)\ge1$. This furthermore forces $f(x)\ge1, \forall x\ge1$, thus $f$ is monotonically increasing. Therefore since $f$ is multiplicative and monotonically increasing we obtain $f(x)=x^c$, plugging this into $P(x,y,z)$ we obtain $c=2$, thus $f(x)=x^2$ So, to sum up $\boxed{f\equiv0\text{ and }f(x)=x^2, \forall x\in\mathbb{R}}$ $\blacksquare$.

F10tothepowerof34 wrote: $f(x)\ge1, \forall x\ge1$, thus $f$ is monotonically increasing. Counterexample $$f(x) = \sin(x) + 2$$

tadpoleloop wrote: F10tothepowerof34 wrote: $f(x)\ge1, \forall x\ge1$, thus $f$ is monotonically increasing. Counterexample $$f(x) = \sin(x) + 2$$ Monotonically increasing comes from multiplicativity and this property, since if $x>y$ but $f(x)<f(y)$ then $f(x/y)<1$

$f\equiv0$ is a solution so assume $f\not\equiv0$ Let $P(x,y,z)$ be the assertion of $(f(x)+1)(f(y)+f(z))=f(xy+z)+f(xz-y)$ $P(0,0,0)\implies (f(0)+1)\cdot 2f(0)=2f(0)\implies f(0)=0$ $P(0,y,z)\implies f(y)+f(z)=f(z)+f(-y)\implies f(y)=f(-y)\implies f$ is an even function $P(x,0,z)\implies (f(x)+1)f(z)=f(z)+f(xz)\iff f(x)f(z)=f(xz)$ which is an equation of the Cauchy type with solution $f(x)=x^c$ for any c *Edit-Proof for boundedness so we can apply Cauchy's functional equation: for $x\geq0, f(x)\geq 0$ $f(x)=f(\sqrt{x})^2\geq0$ Since f is an even function then c is even. c is also positive because $f(0)=0$ $c=2$ is a solution Assume $c\geq4$ $P(1,y,z)\text{ where }y,z\neq0$ $\implies 2(y^c+z^c)=(y+z)^c+(y-z)^c=\sum_{i=0}^{c}\binom{c}{i}(y^iz^{c-i}+y^i(-z)^{c-i})=\sum_{i=0}^{\frac{c}{2}}2\binom{c}{2i}y^{2i}z^{c-2i}=2y^c+2z^c+\sum_{i=1}^{\frac{c}{2}-1}2\binom{c}{2i}y^{2i}z^{c-2i}>2y^c+2z^c$ Contradiction All functions f are $f\equiv0$ and $f(x)=x^2$