gouthamphilomath wrote: $(BUL 2)$ Find all functions $f$ defined for all $x$ that satisfy the condition $xf(y) + yf(x) = (x + y)f(x)f(y),$ for all $x$ and $y.$ Prove that exactly two of them are continuous. Let $P(x,y)$ be the assertion $xf(y)+yf(x)=(x+y)f(x)f(y)$ If $f(0)\ne 0$ $P(x,0)$ $\implies$ $xf(x)=x$ and so $f(x)=1$ $\forall x\ne 0$ and this is indeed a solution, whatever is $f(0)$ If $f(0)=0$ and $\exists u\ne 0$ such that $f(u)=0$, then $P(x,u)$ $\implies$ $f(x)=0$ $\forall x$ If $f(0)=0$ and $f(x)\ne 0$ $\forall x\ne 0$, then $P(x,x)$ $\implies$ $f(x)=1$ $\forall x\ne 0$ And so the solutions : $f(x)=0$ $\forall x$ $f(x)=1$ $\forall x\ne 0$ and $f(0)=a$ And obviously only two of these solutions are continuous : $f(x)=0$ $\forall x$ $f(x)=1$ $\forall x$

Excuse me there's another solution that's not mentioned $ f(x) = 0$ $\forall x \ne 0 $ and $f(0) = a $ It follows from the assertion $ P(x,x) $ Another solution follows from the assertion $ P(x,x) $ which is $ f(x) = 1 $ $\forall x \in S $ and $f(x) = 0 $ $\forall x \in S`-\{0\} $ and $f(0) = k $ for a real constant $ k $, where S is a subset of the set of real numbers that doesn't contain the element 0. but this solution is falsified if $ x \in S $ and $y \in S`-\{0\} $ so another solution would be the one mentioned at the beginning.

Thrax wrote: Excuse me there's another solution that's not mentioned $ f(x) = 0$ $\forall x \ne 0 $ and $f(0) = a $ I'm sorry but if $a\ne 0$ this is not a solution : Choose $x=0$ and $y=1$ so that $f(x)=a$ and $f(y)=0$ $xf(y)+yf(x)=a$ $(x+y)f(x)f(y)=0$ And so we dont have equality.

I'm sorry , you are right, this solution implies that k = 0; Thanks

wrong...

Wrong: 1. $f(x) = f(x)^2$ only when $x \neq 0$, since you divided by $x$. 2. It's possible that $f(x) = 1$ for some $x$ and $f(x) = 0$ for the rest.

Sorry for bumping the Post. But I want to know either my solution is wrong or right. $P(x,y)$ be the assertion. $P(x,1)$: $xf(1)+f(x)=(x+1)f(x)f(1)$ $xf(1)+f(x)=xf(x)f(1)+f(x)f(1)$ $f(x)[f(1)+xf(1)-1]=xf(1)$ Suppose $f(1)=c$ Then $f(x)=\frac{xc}{xc+c-1}$ Plugging this to the original equation we get c=0 and c=1 So,$f(x)=0$ And $f(x)=1$

Solved with vsamc. Let $P(x,y)$ denote the given assertion. $P(0,x): xf(0)=xf(0)f(x)$. If $f(0)\ne 0$, then $f(x)=1$ for $x\ne 0$. This gives the solution \[\boxed{\begin{cases} f(x)=a & \text{ if }x=0 \\ f(x)=1 & \text{ if } x\ne 0 \end{cases}},\]where $a$ is a nonzero constant, which works. Now suppose $f(0)=0$. $P(x,-x): xf(-x)-xf(x)=0\implies xf(-x)=xf(x)$, so $f$ is even. $P(x,-y): xf(y)-yf(x)=(x-y)f(x)f(y)$. Add this with the original FE. We get $2xf(y)=2xf(x)f(y)$. If $f$ is nonzero set $y=k$ with $f(k)\ne 0$. We have $2x=2xf(x)$. So for $x\ne 0$, we have $f(x)=1$ and $f(0)=0$. So $\boxed{\begin{cases} f(x)=0 \text{ if }x=0\\ f(x)=1 \text{ if }x\ne 0 \end{cases}}$, which works. Otherwise, $\boxed{f\equiv 0}$, which also works. Clearly the only continuous solutions are $\boxed{f\equiv 0}$ and $\boxed{f\equiv 1}$, so there are exactly two.

The only such functions are $\boxed{f(x)=0}$ and $\boxed{f(x)=\begin{cases}c&\text{if }x=0,\\1&\text{otherwise.}\end{cases}}$, which clearly work. Consequently, the only two continuous solutions are $f(x)=0$ and $f(x)=1$. Letting $y=1$, we have that $xf(1)+f(x)=(x+1)f(x)f(1)\text{ }(\ast)$. Plugging in $x=1$ yields $2f(1)=2f(1)^2$, which implies that $f(1)=0$ or $f(1)=1$. If $f(1)=0$, in $f(\ast)$ we have that $f(x)=0$. On the other hand, if $f(1)=1$, in $f(\ast)$ we have that $x+f(x)=(x+1)f(x)$, or, equivalently, $xf(x)=x$. When $x\neq0$ we require that $f(x)=1$, but $x=0$ allows $f(x)$ to take on any constant, which yields the second set of solutions above. $\blacksquare$