parmenides51 wrote: Find all non-constant functions $f : Q^+ \to Q^+$ satisfying the equation $$f(ab + bc + ca) =f(a)f(b) +f(b)f(c)+f(c)f(a)$$for all $a, b,c \in Q^+$ . Let $P(x,y,z)$ be the assertion $f(xy+yz+zx)=f(x)f(y)+f(y)f(z)+f(z)f(x)$ Let $c=f(1)$ 1) $c\in\{\frac 13,1\}$

2) If $c=\frac 13$, $fx)=\frac 13\quad\forall x\in\mathbb Q^+$ :

3) If $c=1$ solution $f(x)=x$ :

And so $\boxed{f(x)=x\quad\forall x\in\mathbb Q^+}$, whih indeed fits.

This is my solution: Let $P(a;b;c)$ be the given assertion. Consider $P(a;b;c)$ for $b+c=1$, this gives us that $f(a+bc)=(f(b)+f(c))f(a)+f(b)f(c)$ I now claim that either the function is constant either $f(b)+f(c)=1$, assume that none of these happens, let $f(b)+f(c)=k_1$ and $f(b)f(c)=k_2$, Letting $bc=\frac{x}{y}$ and iterating it $y$ times, we get $f(a+x)=k_1^yf(a)+$ some constant. Denote that constant by $m$ and let $k_1^y$ be $k$, this rewrites as $f(a+x)=kf(a)+m$ if $k$ would be 1, the claim would be proven, thus assume it is not, now we get $f(a+nx)=k^nf(a)+m\frac{k^n-1}{k-1}$, and by placing $a$ to be precisely $x$ we get $f(nx)=c_1k^n+c_2$ for constant $c_1$ and $c_2$, and taking $P(nx;nx;nx)$ for natural $n$ we see that $c_1k^{3n^2x}+c_2=3(c_1k^n+c_2)^2$, and now if $k>1$, for large enough $n$, $LHS>>RHS$, if $k<1$ again for $n \rightarrow \infty$, we get $3c_2^2=c_2$, and if $c_2=0$ we are instantly done, else subtract $\frac{1}{3}$ from both sides and divide by $k^n$ to conclude again. Now if the function is constant obviously it is $\frac{1}{3}$, Else we know that as $\frac{1}{2}+\frac{1}{2}=1$, $f(\frac{1}{2})+f(\frac{1}{2})=1$, which from $P(x;\frac{1}{2};\frac{1}{2})$ means that $f(x+\frac{1}{4})=f(x)+\frac{1}{4}$, or that $f(x+1)=f(x)+1$, moreover taking $b+c=1$ again, we see that $f(x+bc)=f(x)+f(b)f(c)$, meaning after iterating that$f(b)f(c)=bc$, and together with $f(b)+f(c)=1$, that $f(x) \in $ {$x;1-x$} for any $x<1$, and now assuming there exists some $x$ with $f(x)=1-x$ and $x\le \frac{1}{4}$ we get a fast contradiction from $P(x;x;x)$, and together with $f(x+\frac{1}{4})=f(x)+\frac{1}{4}$, we conclude that $f(x)=x \; \forall \; x \in \mathbb{Q_+}$ Everything above does not look completely natural, however for easyer comprehension try and think in the following way: After substituting $b+c=1$ we get something of the form $f(x+c)=af(x)+b$, which after iterations shows that unless $a=1$, the function is exponential which intuitively can not be true, and after proving this we are kind of done.