if $f$ is constant then $ f=0 $ or $f=1/2$ Assume that $f$ isn't constant. Let $P(x, y) $ be the assertion of : $f(xy+xf(x)) =f(x) (f(x) +f(y)) $ $P(0,y) \Rightarrow f(0)=0$ suppose that $f(a) =0$ $P(a, y) \Rightarrow a=0$. $P(x, - f(x)) \Rightarrow f(x)=-f(-f(x)) $. $P(-f(x), x) \Rightarrow f(-xf(x) +f(x)^2)=0 $ $\Rightarrow f(x) =x $

Let $P(x,y)$ be the equation. The only constant functions that work are $f\equiv 0,1/2$. Henceforth assume $f$ is nonconstant. We will show that $f(x)=x$ is the only other solution; it clearly works. Lemma: [Weak injectivity] $f(a)=0 \implies a=0$. Proof: Suppose $f(a)=0$ for some $a\not=0$. Then $P(a,y)$ gives $f(ay)=0$. Varying $y$ over $\mathbb{R}$ then gives $f\equiv 0$, contradiction. So $a=0$, as claimed. $\square$ Claim: $f(-f(x)) = -f(x)$, for any $x\not = 0$. Proof: $P(0,y)$ gives $f(0)=f(0)[f(0)+f(y)]$, so $f(0)=0$ or $f$ is constant, which is impossible. Hence $f(0)=0$. Then $P(x,-f(x))$ gives $0=f(x)[f(x) + f(-f(x))]$, so $f(x)=0$ or $f(-f(x)) = -f(x)$ for each $x$. But $x\not =0$, so $f(x)\not =0$ by the lemma. Hence $f(-f(x)) = -f(x)$, as claimed. $\square$ Now, $P(-f(y),y)$ gives $f(-yf(y) - f(y)f(-f(y)))=0$. So the lemma tells us that $-yf(y) - f(y)f(-f(y))=0$, i.e. $-yf(y)+f(y)^2=0$. So if $y\not=0$, then $f(y)\not = 0$, so cancelling gives $f(y)=y$. And we already showed $f(0)=0$. Therefore, $f(x)=x$ for all $x$. This completes the proof.

Remarks

USJL wrote: Let $\mathbb{R}$ denote the set of all real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$, \[f(xy+xf(x))=f(x)\left(f(x)+f(y)\right).\] Let $p(x,y)$ be the assertions for \[f(xy+xf(x))=f(x)\left(f(x)+f(y)\right).\]From $p(0.0)$ we get that $f(0)=0$ or $f(0)=1\2$ $Case.1.1)$ $f(0)=1\2$ Taking $p(0,x)$ we get that $f(x)=1\2$ $Case .1.2)$ $f(0)=0$ taking $p(x,0)$ we get that $f(xf(x))=f(x)^2$ $p(x,-f(x))$ gives us that $0=f(x)(f(x)+f(-f(x)))$ Which means that we have two cases $Case 2.1$ $f(x)=0$, which works and $Case 2.2$ $f(x)=-f(-f(x))$,which means that $f(x)^2=f(-f(x))$ ...(1) From $p(1,x)$ we get $f(x+f(1))=f(f(1))+xf(1)$- surjective so $\exists a\in\mathbb R$ such that $f(a)=-1$ Now using ...(1) ,$x-->a$ we get that $1=f(1)$ Now from $p(1,x)$ we get that $f(x+1)=x+1$ And now putting $p(x+1,-x)$ we get that $f(x) = x \forall x \in \mathbb{R}$

2020 Taiwan TST Round 2 Quiz 2 P1 wrote: Let $\mathbb{R}$ denote the set of all real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$, \[f(xy+xf(x))=f(x)\left(f(x)+f(y)\right).\] If $f$ is constant then $f \equiv 0 $ and $f \equiv 1/2$ are the only solutions. From now on, assume that $f$ is non-constant. Let $P(x,y)$ be the given assertion. To start, $P(0,x)$ implies $f(0)=0$ or $f(x)+f(0)=1$. Since $f$ is non-constant, we have $f(0)=0$. Hence, $P(x,0) \Rightarrow f(xf(x))=f^2(x), \, (*)$. Claim 1: $f(x)=0$ if and only if $x=0$. Proof: Let $f(k)=0$ with $k \neq 0$. Then $P(k,x)$ implies $f(kx)=0$ hence $f \equiv 0$, contradiction $\blacksquare$. Claim 2: $f(x+1)=f(x)+1$. Proof: $P(x,-f(x))$ implies $f(x)(f(x)+f(-f(x)))=0$. Since $f(k)=0 \Rightarrow f(k)+f(-f(k))=0$, we conclude that $f(-f(x))=-f(x)$. Take $x \rightarrow -f(x)$ here to obtain $f(f(x))=f(x)$. Thus, $f(f(1))=f(1)$ and $f(f(1))=f^2(1)$ due to $(*)$. Therefore $f(1)=1$. Hence, $P(1,x)$ gives the desired $\blacksquare$ Claim 3: $f$ is odd. Proof: Using Claim 2, we easily obtain $f(-1)=-1$, hence $P(-1,-x) \Rightarrow f(x+1)=1-f(-x) \Rightarrow f(-x)=-f(x)$, as desired $\blacksquare$ To conclude, $P(x,-x) \Rightarrow f(xf(x)-x^2)=0$, hence by Claim 1 $xf(x)-x^2=0 \Rightarrow f(x)=x$, which evidently satisfies. Thus, only solutions are $f \equiv 0 , f \equiv 1/2$ and $f \equiv \rm id$.

\[P(0,y) \implies f(0)=f(0)(f(y)+f(0))\].unless $f(0)=0$ we have $f(x)=c$ substituting this in the equation we get $f(x)=0$ or $f(x)=1/2$. So,now we assume $f(0)=0$. Now,assume there exist some $u \not =0$ for which $f(0)=0$ \[P(u,y) \implies f(uy)=0 \implies f(y)=0 \forall y\].so,now we can assume that $f(u)=0 \implies u=0$. Now,\[P(x,-f(x)) \implies 0=f(x)(f(x)+f(-f(x)) \implies f(-f(x))=-f(x)\]\[P(-f(x),x) \implies f(-f(x)x-f(x)f(-f(x)))=0 \implies f(x)^2=f(x)x \implies f(x)=x\]

The only constant solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=\frac12}$. Assume now that $f$ is nonconstant. Claim: $f(k)=0\Leftrightarrow k=0$ If $\exists k:f(k)=0$, then $P(k,y)\Rightarrow f(ky)=0$, and since $f$ is nonconstant, $k=0$. $P(0,y)\Rightarrow f(0)=f(0)(f(0)+f(y))$ and since $f$ is nonconstant, $f(0)=0$ $P(x,-f(x))\Rightarrow 0=f(x)(f(x)+f(-f(x)))$ Let $x\ne0$, then $f(-f(x))=-f(x)$ (this actually holds $\forall x$) due to our claim. Switching things around a bit, we obtain: $P(-f(x),x)\Rightarrow f(-xf(x)+f(x)^2)=0$, so $-xf(x)+f(x)^2=0$ and thus $\boxed{f(x)=x}$ because of the claim.

Was i the only person who went "ok Cauchy done" $P(x,y): f(xy+xf(x))=f(x)^2+f(x)f(y)$. $P(0,y): f(0)=f(0)^2+f(0)f(y)$. Hence, either $f$ is constant or $f(0)=0$. (the only constant solutions are $0$ and $\frac{1}{2}$) Assume $f$ is non-constant Note that $f$ is injective at $0$. This is because if $f(c)=0$ then $P(c,y): f(cy)=0$. $P(-1,-f(-1)): f(-1)^2+f(-1)f(-f(-1))=0\implies f(-f(-1))=-f(-1)$. $P(-1,0): f(-f(-1))=f(-1)^2$. Hence, this implies $f(-1)=-1$. Then, $f(1)=f(-f(-1))=-f(-1)=1$. This means $P(1,x): f(x+1)=f(x)+1$. Thus, $P(x,y+1): f(xy+x+xf(x))=f(x)^2+f(x)f(y)+f(y)=f(xy+xf(x))+f(x)$. This simplifies to $f(x+y)=f(x)+f(y)$. Hence, the original condition becomes $f(xy)+f(xf(x))=f(x)^2+f(x)f(y)$. From $P(x,0)$, we get $f(xf(x))=f(x)^2$, hence $f(xy)=f(x)f(y)$. Since $f$ is multiplicative and additive, $f\equiv x$, done. Why yall so smart with the $P(-f(x),x)$

If $f$ is constant then $f(x)\in \{0,.5\}~ \forall x,$ this works. So assume that $f$ is not constant. Denote the assertion by $P(x,y)$ and note that $f(0)=0.$ Let $f(x_0)=0,$ then $P(x_0,x)$ forces $x_0=0.$ Then $P(x,-f(x))$ yields $-f(x)=f(-f(x))$ for all $x\neq 0.$ Finally $P(-f(x),x)$ yields $f(x)=x,$ which fits.

gghx wrote: Why yall so smart with the $P(-f(x),x)$ Sorry for double post It makes the RHS vanish so that we may conduct injectivity at zero.

The only solutions are $\boxed{f\equiv 0, f\equiv \frac{1}{2}, f(x) = x}$, which work. Let $P(x,y)$ denote the given assertion. $P(0,0): f(0)\in \left\{0,\frac{1}{2}\right\}$. $P(0,x): f(0) = f(0)(f(x) + f(0))$. If $f(0) = \frac{1}{2}$, then obviously $f(x) = \frac{1}{2}$ for all $x$. Now assume $f(0) = 0$. The only constant solution with $f(0) = 0$ is $f\equiv 0$, so assume $f$ is nonconstant. Claim: $f$ is injective at $0$. Proof. Suppose $f(a) = 0$ and $a\ne 0$. $P(a,x): f(ax) = 0$, so $f\equiv 0$, contradiction. $\square$ $P(x,-f(x)): f(x)(f(x) + f(-f(x))) = 0$, so $f(-f(x)) = -f(x)$ for all $x\ne 0$. $P(-f(x),x): f(-xf(x) +f(x)^2) = 0$, so $-xf(x) + f(x)^2=0$, which means for all $x\ne 0$, $f(x) = x$.

The only solutions are $f(x)=x$ and $f(x)\equiv c$ for $c\in \left \{ 0,\frac 12 \right \}$ (which clearly work). If $f$ is a constant $c$ we obtain $2c^2-c=0,$ which gives the second case. Next we suppose the opposite. Substituting $x=0$ we get $f(0)(f(y)+f(0)-1)=0\implies f(0)=0.$ Substituting any $x=u:f(u)=0$ we obtain $f(uy)=0\implies u=0$. Then substituting $y=-f(x)$ we obtain $0=f(x)-f(-f(x)),$ so substituting $x=-f(y)$ we finally get $f(-yf(y)-f(y)^2)=0\implies f(y)=y.$

Let $P(x,y)$ be the assertion. $P(0,x):f(0)=f(0)(f(0)+f(y))$ So either $f(0)=0$ or $f(y)\equiv c$ If $f(y)\equiv c$, plug in the FE and we get $c\in\{0,\frac 1 2 \}$ Let $a$ be a number such that $f(a)=0$. $P(a,y):f(ay)=0\Rightarrow a = 0$ If $f(0) = 0$: $P(x,0):f(xf(x))=f(x)^2$ $P(x,-f(x)):f(x)(f(x)+f(-f(x))=0\Rightarrow f(x)=-f(-f(x))$ $P(-f(x),x):f(-xf(x)+f(x)^2)=f(-f(x))(f(-f(x))+f(x))=0$ So $-xf(x)+f(x)^2=0$, which implies $f(x)=x$, done.

\[f(xy+xf(x))=f(x)\left(f(x)+f(y)\right)\]Answers are $f\equiv 0,f\equiv \frac{1}{2},f(x)=x$. Suppose that $f$ is non-constant. Let $P(x,y)$ be the assertion. $P(0,y)$ yields $f(0)(f(0)-1+f(y))=0$. Since $f$ is non-constant, $f(0)=0$. Also note that $f(a)=0$ implies $f(ay)=0$ hence $a=0$. \[P(x,1-f(x)): \ \ f(x)=f(x)(f(x)+f(1-f(x)))\overset{f(x)\neq 0}{\iff} f(1-f(x))=1-f(x)\]For $x\neq 0$. Also $P(x,0)$ implies $f(xf(x))=f(x)^2$. We have $f(xy)=f(x)(f(x)+f(y-f(x)))$ Plugging $xf(x),y$ and $x,yf(x)$ gives \[f(x)(f(x)+f((y-1)f(x)))=f(xyf(x))=f(x)^2(f(x)^2+f(y-f(x)^2)\]For $x\neq 0$, choose $y=1$ to observe that $f(x)^2+f(1-f(x)^2)=1\iff f(1-f(x)^2)=1-f(x)^2$. $P(1-f(x)^2,y)$ implies \[f(y-yf(x)^2)=(1-f(x)^2)(1-f(x)^2+f(y-1+f(x)^2))\]For $x=0$, we see that $f(y)=1+f(y-1)$ thus, $f(x+n)=f(x)+n$ for all integers $n$. Compare $P(x,y)$ with $P(x,y+n)$ to get \[f(xy+xn)=f(x)^2+f(x)f(y+n-f(x))=f(x)^2+f(x)f(y-f(x))+nf(x)\]Hence $f(xy+xn)=nf(x)+f(xy)$. We can write $y$ instead of $xy$. So $f(y+xn)=f(y)+nf(x)$. Plugging $y=0$ yields $f(xn)=nf(x)$ thus, $f(y+xn)=f(y)+nf(x)=f(y)+f(xn)$. For $\frac{x}{n},y$ we get that $f(x+y)=f(x)+f(y)$ subsequently, $f$ is additive. \[f(xy)=f(x)(f(x)+f(y-f(x))=f(x)f(x-f(x)+y)\]Choosing $y=1$ yields \[f(x)=f(x)f(x-f(x)+1)\iff 1=f(x-f(x)+1)=f(x-f(x))+1\iff f(x-f(x))=0\iff f(x)=x\]As desired.$\blacksquare$

Yes, simple as a pimple. The answers are $f(x) = 0$ for all $x\in \mathbb{R}$ , $f(x)= \frac{1}{2}$ for all $x\in \mathbb{R}$ and $f(x)=x$ for all $x \in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(xy+xf(x))=f(x)\left(f(x)+f(y)\right)$ for real numbers $x$ and $y$. We start off with some easy observations. Note that from $f(0,0)$ we have that, \[f(0)=2f(0)^2\]Thus, $f(0)=0$ or $f(0)=\frac{1}{2}$. If $f(0)=\frac{1}{2}$ then $P(0,x)$ implies that \[\frac{1}{2}=f(0)=f(0)(f(0)+f(x))=\frac{1}{4} + \frac{f(x)}{2} \]So, $f(x) = \frac{1}{2}$ for all real numbers $x$. Thus, in what follows we consider $f(0)=0$, and that $f$ is not constant zero. Claim : The function $f$ is injective at zero, i.e $f(x)=0$ if and only if $x=0$. Proof : One of the directions is already done. For the other note that, if there exists $\alpha \ne 0$ such that $f(\alpha)=0$ then, $P(\alpha,x)$ gives \[f(\alpha x)=f(\alpha x+\alpha f(\alpha))=f(\alpha)(f(\alpha)+f(x))=0\]Replacing $x \to \frac{t}{\alpha}$, we have that $f(t)=0$ for all $t\in \mathbb{R}$, which is a clear contradiction. Thus, there exists no such $\alpha$ proving the claim. Now, from $P(x,-f(x))$ (for $x\ne 0$) we have, \[0=f(0)=f(-xf(x)+xf(x))=f(x)(f(x)+f(-f(x)))\]Thus, $f(x)(f(x)+f(-f(x)))=0$ which as a result of our previous claim implies $f(-f(x))=-f(x)$ for all $x\ne 0$. Now, looking at $P(-f(x),f(x))$ we have, \[0=f(-f(x)^2+f(x)^2)=f(-f(x)^2 + -f(x)f(-f(x)))=f(-f(x))(f(-f(x))+f(f(x)))=-f(x)(f(f(x))-f(x))\]which implies that $f(f(x))=f(x)$ for all $x\ne 0$. Finally, from $P(f(x),-x)$ we have, \[0=f(-xf(x)+xf(x))=f(-xf(x)+f(x)f(f(x)))=f(f(x))(f(f(x))+f(-x))=f(x)(f(x)+f(-x))=f(-x^2+xf(x))\]which since $f$ is injective at zero implies, $xf(f)=x^2$ for all $x\ne 0$. Thus, $f(x)=x$ for all $x\ne 0$, which combined with $f(0)=0$ implies $f(x)=x$ for all $x\in \mathbb{R}$. Thus, all the solutions to the given functional equation are of the claimed forms, and we are done.