parmenides51 wrote: Find all functions $f : R \to R$ satisfying $(x^2 + y^2)f(xy) = f(x)f(y)f(x^2 + y^2)$ for all real numbers $x$ and $y$. Posted (and solved) many many times (at least 2013, 2014, twice in 2016, twice in 2017). Dont hesitate to use the search function (see here ). Set (for example copy/paste) in the "search term" field the exact following string : +"(x^2+y^2)f(xy)" +"f(x)f(y)f(x^2+y^2)" You'll get in the ten first results (excluded your own post and this post itself) all the help you are requesting for.

so for fuctionals you search without spaces, thanks just giving one link with the 5 functions that are solutions of this one, for the post collection

Posting for storage Let $P(x,y):=(x^2+y^2)f(xy)=f(x)f(y)f(x^2+y^2)$ $P(0,0)$ yields $f(0)^3=0\Longrightarrow f(0)=0$ Now, assume that $\exists\alpha\text{ such that }f(\alpha)=0\text{ and }\alpha\neq0$ $P(x/\alpha,\alpha)$ yields $(x^2/\alpha^2+\alpha^2)f(x)=0$ thus either $f\equiv0, \forall x\in\mathbb{R}$ or $x^2/\alpha^2=-\alpha^2$, however the latter forces $x^2=-\alpha^4\Longrightarrow -\alpha^4\ge0\Longrightarrow\alpha=0$ Which is clearly a contradiction. Thus $f\equiv0$ is one solution So from now on assume that $f\not\equiv0$ Furthermore letting $x>1$ and by $P(\sqrt{x-1},1)$ we obtain $x=f(1)f(x)\Longrightarrow f(x)=\frac{x}{c}\text{ where }c=f(1)$ however this is also true for $x=1$ thus $f(x)=x, \forall x\ge1$ Plugging this $P(x,y)$ we obtain that that $c=\pm1$ however for now let $c=1$ $P(x,1/x)$ yields $(x^2+1/x^2)=f(x)f(1)f(x^2+1/x^2)=f(x)f(1)(x^2+1/x^2)\Longrightarrow f(x)f(1/x)=1$ furthermore this forces $f(x)=x, \forall x\ge0$ Inspecting $P(x,-1)$ yields $f(x)f(-1)=f(-x)$ thus $f(x)=x\text{ and }f(x)=|x|, \forall x\in\mathbb{R}$ However notice that the original $\text{FE}$ is symmetric thus $f(x)=-x\text{ and }f(x)=-|x|, \forall x\in\mathbb{R}$ In conclusion $\boxed{f(x)=\pm x\text{ and }f(x)=\pm |x|, \forall x\in\mathbb{R}}$ $\blacksquare$.