wanwan4343 wrote: Find all functions $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$ such that \[(f(x))^2f(2y)+(f(y))^2f(2x)=2f(x)f(y)f(x+y)\] for all $x,y\in\mathbb{Q}$ Let $P(x,y)$ be the assertion $f(x)^2f(2y)+(f(y)^2f(2x)=2f(x)f(y)f(x+y)$ $f(x)$ solution implies $cf(x)$ solution and so WLOG $f(0)=1$ (since $0\notin f(\mathbb Q)$) $P(x,0)$ $\implies$ $f(2x)=f(x)^2$ and $P(x,y)$ becomes $f(x+y)=f(x)f(y)$ Hence the solution : $\boxed{f(x)=b e^{ax}}$ $\forall x\in\mathbb Q$, which indeed is a solution, whatever is $a\in\mathbb R$, $b\in\mathbb R\setminus\{0\}$

What about non continuous solutions?

WolfusA wrote: What about non continuous solutions? What is your definition of continuous / non continuous function when domain is $\mathbb Q$ ???????

$f:\mathbb Q\to \mathbb R$ is continuous on set $\mathbb Q $ iff for every sequence $(x_n)$ where $\forall n \quad x_n\in \mathbb Q$ and $\lim_{n\to \infty}x_n=y$ holds $\lim_{x_n\to y} f(x_n)=f(y)$

WolfusA wrote: What about non continuous solutions? There's no need to consider it, isn't it? Because the domain is rational number, then we can directly apply Cauchy Functional equation as pco did.

WolfusA wrote: $f:\mathbb Q\to \mathbb R$ is continuous on set $\mathbb Q $ iff for every sequence $(x_n)$ where $\forall n \quad x_n\in \mathbb Q$ and $\lim_{n\to \infty}x_n=y$ holds $\lim_{x_n\to y} f(x_n)=f(y)$ Huhhh ! And what about $y\notin\mathbb Q$ so that $f(y)$ is undefined ?

I see your point. So what is the criterium then for using Cauchy's functional equation, because I thought it's that function is continuous.

Given that the domain is $\mathbb{Q}$, you just apply Cauchy's functional equation the normal way on $\log{f(x)}+\log{f(y)}=\log{f(x+y)}$.

This solutions also works. $\forall x\in\mathbb Q$ $f(x)=bs^{ax}$ whatever is $a\in\mathbb R$, $s,b\in\mathbb R\setminus\{0\}$,

InCtrl wrote: Given that the domain is $\mathbb{Q}$, you just apply Cauchy's functional equation the normal way on $\log{f(x)}+\log{f(y)}=\log{f(x+y)}$. You can't do that, because $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$, and you can't take a logarithm of negative number.

WolfusA wrote: InCtrl wrote: Given that the domain is $\mathbb{Q}$, you just apply Cauchy's functional equation the normal way on $\log{f(x)}+\log{f(y)}=\log{f(x+y)}$. You can't do that, because $f:\mathbb{Q}\rightarrow\mathbb{R} \setminus \{ 0 \}$, and you can't take a logarithm of negative number. Actually, f(2x)=f(x)^2, so nothing would be negative.