Are Russia TST's being shared anymore?

@below hmmmmmm, plugging back is hard sometimes

Huh? Actually, easy to check that $f(x)=2-x$ also works...

Another argument in the case $f(0) \ne 0$: $P(0,x)$ shows that $f(f(0))=0$. Now $P(f(0),0)$ shows that $2f(0)=f(0)^2$. Hence, if $f(0) \ne 0$, we must have $f(0)=2$. But then $P(x,0)$ shows that $f$ is injective and $P(x,2)$ shows that $f(2x+f(x))=f(x+2)$, hence $f(x)=2-x$ by injectivity.

oVlad wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying \[f(xy+f(x))+f(y)=xf(y)+f(x+y),\]for all real numbers $x,y$. Let $P(x,y)$ denote $f(xy+f(x))+f(y)=xf(y)+f(x+y)$ $P(0,0)$ gives $f(f(0))=0$ $P(1,0)$ gives $f(f(1))=1$ Claim: The function is injective when $f(0)\neq 0$

Case 1:$f(0)\neq0$

Case 2:$f(0)=0$

Assuming there exists a satisfying function, the symbol $P(u,v)$ is the substitution of $x$ by $u$, $y$ by $v$ into the function equation$$f(xy+f(x) )+f(y)=xf(y)+f(x+y)$$ $P(0,0)\implies f(f(0))=0$ $P(f(0),0)\implies 2f(0)=f(0)^2$ so $f(0)=0$ or $f(0)=2$ $\textit{Case 1.}$ $f(0)=0$ $\bullet$ Considering $f(1) \neq 1$ then $P(1,y)\implies f(y+f(1))=f(y+1)$ Put $T=f(1)-1 \neq 0$ then we have $f(y)=f(y+T),\forall y \in \mathbb{R}$ From $P(x,y+T)$ and combining with the original functional equation we get $f(x(y+T)+f(x))=f(xy+f(x))$ Here for every $x \neq 0$, replace $y$ by $\dfrac{-f(x)}{x}$, we get $f(Tx)=0,\forall x \neq 0$ That is, $f(x)=0,\forall x \neq 0$, combined with $f(0)=0$ we get the function $\boxed{f(x)=0,\forall x \in \mathbb {R}}$ is a solution $\bullet$ Consider $f(1) = 1$ Assuming there exists a number $b \neq1$ so that $f(b)=1$ then from $P(b,1)$ we get something absurd so $f(x)=1\Leftrightarrow x=1$ Assuming there exists $a \neq 0$ so that $f(a)=0$ then from $P(a,1)$ we have $f(a+1)=1-a$ $P(a+1,0)\implies f(1-a)=1-a$ $P(1-a,a)\implies f(1-a^2)=1$ deduce $1-a^2=1$ so $a=0$, contradiction so $f(x)=0\Leftrightarrow x=0$ For each $x \neq 1$ then $P(x,\dfrac{x-f(x)}{x-1})\implies f(\dfrac{x-f(x)}{x-1})=0,\forall x \neq 1$ That $f$ is unique at $0$ so $f(x)=x,\forall x \neq 1$, combined with $f(1)=1$ we have $\boxed{f(x)=x, \forall x \in \mathbb{R}}$ $\textit{Case 2.}$ $f(0)=2$ $P(0,y)\implies f(2)=0$ $P(x,0)\implies f(f(x))=2x-2+f(x)$, from here we also have $f(x)=0\Leftrightarrow x=2$ For each $x \neq 1$ then $P(x,\dfrac{x-f(x)}{x-1})\implies f(\dfrac{x-f(x)}{x-1})=0,\forall x \neq 1$ Hence $f(x)=2-x,\forall x \neq 1$ $P(1,1)\implies f(1+f(1))=0\Rightarrow f(1)=1$ In short, we get $\boxed{f(x)=2-x,\forall x \in \mathbb{R}}$

let $P(x, y)$ denote the given assertion $P(1, y)$ implies $f(y + f(1)) = f(y + 1)$ so if $f(1) \neq 1$ then $f$ is periodic with period $p = f(1) - 1$; hence $P(x, y + p) - P(x, y)$ implies that $f(xy + xp + f(x)) = f(xy + f(x))$ but taking $x = \frac{u - v}p, y = \frac{v - f(x)}x$ gives $f(u) = f(v)$ so $f$ is constant; checking $f(x) = c$ implies $2c = (x + 1)c$; $x = 2 => c = 0$, so $f(x) = 0$ is the only solution in this case; hence assume $f(1) = 1$. $P\left(x, \frac{x - f(x)}{x - 1}\right)$ gives $xy + f(x) = x + y$ so that $(x - 1)f(y) = 0$; hence for all $x \neq 1$ we have $f\left(\frac{x - f(x)}{x - 1}\right) = 0$. Then $P(x, 0)$ implies $f(f(x)) - f(x) = (x - 1)f(0)$; $x = \frac{t - f(t)}{t - 1}$ implies $\frac{2 - t}{t - 1} f(0) = 0$ which means that either $f(t) = 2 - t$ for all $t \neq 1$ (which combined with $f(1) = 1$ implies $f(x) = 2 - x$ for all $x$, or taking some $t$ for which $f(t) \neq 2 - t$ we get $f(0) = 0$. Hence assume that $f(0) = 0$ as well. Assume we have $f(a) = 1$ for some $a$; then $P(a, 1)$ implies $(a - 1)f(1) = 0$ so that $a = 1$ and so $f$ is injective at 1. Now assume we have $f(a) = 0$ for some $a$ instead. $P(a, 1)$ implies $f(a + 1) = 1 - a$, so $P(a + 1, 0)$ implies $f(1 - a) = 1 - a$ and finally $P(1 - a, a)$ implies $f(1 - a^2) = f(1)$ which by the previous implies $a^2 = 0 \implies a = 0$ and so $f$ is injective at 0. But since we got $f\left(\frac{x - f(x)}{x - 1}\right) = 0$ for all $x \neq 1$ we hence have $\frac{x - f(x)}{x - 1} = 0$ for all $x \neq 1$, which implies $f(x) = x$ for all $x \neq 1$; combined with $f(1) = 1$ we get that $f(x) = x$. Hence the only solutions are $f(x) = x, f(x) = 0, f(x) = 2 - x$ and these all work.

Nice one! Let $P(x,y)$ be the assertion. We begin by proving some claims. Claim 1 : $f\left(\frac{x-f(x)}{x-1}\right) = 0 $ for all $x \neq 1$. Proof : Ley $x\neq1$ and consider $y=\frac{x-f(x)}{x-1}$ so that $xy+f(x)=x+y$. This gives $f(y)=xf(y)$, and as $x\neq 1$, $f(y)=0$. $\square$ Claim 2 : If $f(x)=f(y)\neq0$, then $x=y$. Proof : Assume $f(x)=f(y)\neq0$. Compare $P(x,y)$ and $P(y,x)$, we have $xf(y)=yf(x)$. Hence $x=y$. $\square$ Claim 3 : $f(0)=0$ or $2$. Proof : $P(0,0) : f(f(0))=0$. Hence, $f(f(f(0)))=f(0)$. $P(f(0),0) : f(0) + f(0) = f(0)\cdot f(0) + 0$. Hence $f(0)=0$ or $2$. $\square$ Case 1 : $f(0)=2$. $P(x,0) : f(f(x)) + 2 = 2x + f(x) \implies f$ is completely injective. $P(0,0) : f(f(0))=0 \implies f(2)=0$. From claim 1 for all $x\neq 1$, $f\left(\frac{x-f(x)}{x-1}\right) = 0 = f(2) \implies \frac{x-f(x)}{x-1} =2 \implies f(x)=2-x$. Hence, for all $x\neq 1$, $f(x)=2-x$. As $f$ is injective, we must have $f(1)=2-1$ and hence $\boxed{f(x)=2-x}$ for all $x$. Case 2 : $f(0)=0$. $P(x,0) : f(f(x))=f(x)$. Hence if $f(x)\neq 0$, claim 2 implies $f(x)=x$. Hence, $f(x) = 0 $ or $x$ for all $x$. $P(1,y) : f(y+f(1))=f(y+1)$. If $f(1)=0$, $f(y)=f(y+1)$. As $f(x)=0$ or $x$, we must have $f(y)=0$ for all $y$, and hence $\boxed{f \equiv 0}$. Otherwise, $f(1)=1$. $P(x-1,1) : f(x-1+f(x-1)) +1 = x-1 + f(x)$. But $f(x-1+f(x-1)) = 0$ or $x-1$ or $x-1+x-1$. Hence, $f(x) = 2-x$ or $1$ or $x$, for all $x$. But we also have $f(x) = 0$ or $x$ for all $x$. Hence, for all $x\neq 2$, we must have $f(x) = x$. Considering $P(e^\pi, 2-e^\pi) $ and using $f(x)=x$ for all $x\neq 2$, we get $f(2)=2$. Hence, $\boxed{f(x)=x}$ for all $x$. Verifying $f(x)=x$, $f(x)=2-x$ and $f\equiv 0$ all work and hence we are done!

The answer is $f \equiv 0$, $f(x)=x$, and $f(x)=2-x$, which clearly work. Let $P(x,y)$ denote the assertion. First suppose that $f(a)=f(b)=k \neq 0$. Then by comparing $P(a,b)$ and $P(b,a)$ we find that $ak=bk \implies a=b$, hence $f$ is injective over inputs that don't get mapped to zero. From $P(1,y)$ we find that $f(y+f(1))=f(y+1)$. If there exists some value of $y$ such that $f(y+1) \neq 0$, then this implies $f(1)=1$; otherwise, $f \equiv 0$, so suppose the former. The key claim is that in this case, there is at most one value of $x$ such that $f(x)=0$. First, by comparing $P(a,0)$ and $P(b,0)$, we find that $f$ is either injective everywhere, or $f(0)=0$. If the former is true, the claim is immediately done, so suppose that $f(0)=0$. Then $P(x,0)$ implies $f(f(x))=f(x)$ for all $x$, i.e. $f$ is the identity on its image. Suppose that there exists some $a \neq 0$ such that $f(a)=0$. Then from $P(a,1)$ we obtain $1=a+f(a+1) \implies f(a+1)=-a+1 \implies f(-a+1)=-a+1$. On the other hand, from $P(-a+1,a)$, we find that $f(-a^2+a+f(-a+1))=1=f(1) \implies f(-a+1)=a^2-a+1$. Comparing this with the previous result implies that $a^2=0$, or $a=0$: contradiction. Now, for $x \neq 1$, $P(x,\tfrac{x-f(x)}{x-1})$ implies $f(\tfrac{x-f(x)}{x-1})=0$, hence $\tfrac{x-f(x)}{x-1}$ equals some constant $c$ for all $x \neq 1$ (specifically, the unique $c$ such that $f(c)=0$), or equivalently $$x-f(x)=cx-c \iff f(x)=(1-c)x+c.$$Note that this is true for $x=1$ as well, since it is known that $f(1)=1$. By plugging in $x=c$ we either have $c=0$ or $c=2$, which yields the two remaining solutions. $\blacksquare$

Let $P(x,y)$ be the assertion: From $P(0,0)$ you get $f(f(0))=0$ Then from $P(f(0),0)$ you get $f(0)(f(0)-2)=0$ This gets us to $3$ cases: $\textcolor{red}{Case 1:}$ $f$ is a constant function. From $P(0,0)$ we know that $f(x)$ can be equal to $0$ or $2$.If we check it we see that $\boxed{f\equiv0}$ is the only solution in this case. $\textcolor{red}{Case 2:}$ $f(0)=0.$ From $P(x,0)$ we get $f(f(x))=f(x)\implies\boxed{f(x)=x}$ is a solution. If we check it we see that it is true. $\textcolor{red}{Case 3:}$ $f(0)=2.$ From $f(f(0))=0$ we get $f(2)=0.$ $\textcolor{blue}{Claim:}$ $f$ is an injective function. $\textcolor{blue}{Proof:}$Let $f(x)=f(y)$ for some integer $y\neq x.$From $P(x,0)$ we get $f(f(x))=2x-2+f(x).$ Then $f(f(x))=f(f(y))=2x-2+f(x)=2y-2+f(x)$ meaning $x$ must be equal to $y.$ Contradiction. Our claim is right. Since $f(2)=0$, from $P(x,2)$ we have $f(2x+f(x))=f(x+2)\stackrel{\text{Injectivity}}{\implies}\boxed{f(x)=2-x}$ as a solution. If we check it we see that it is true. Thus the three solutions are $f(x)=0,f(x)=x,f(x)=2-x.$ And so we are done.

Cute! Clearly we have $f(\frac{f(x)-x}{1-x})=0$, call this equation $(*)$ If $f(0) =0$ then $f(f(x))=f(x)$ so by $P(f(x),x)$ implies $f(x)=0,x$ for every $x$, it is easy to avoid the point wise trap, so we get the solutions $f(x)=x$ and $f(x)=0$ for all $x \in \mathbb{R}$. If $f(0) \neq 0$ then by $P(x,0)$ implies $f$ is injective, so $f(x)$ is linear, which gives $f(x)=2-x$ by using $(*)$ in $P(x,0)$ for all $x \neq 1$, now $P(x,1)$ gives $P(x+f(1))=f(x+1)$ so if $f(1) \neq 1$ then $f$ is periodic, and if there exists a value not equal to $0$ then put $y$ that value and then we get contradiction implies $f \equiv 0$ contradiction! So $f(1)=1$ and the last solution is $f(x)=2-x$

We claim the only functions are $\boxed{f(x)\equiv 0}$, $\boxed{f(x)\equiv x}$, and $\boxed{f(x)\equiv 2-x}$. It is easy to check that these work. Let $P(x,y)$ denote the given assertion. $P(0,0)$ gives $f(f(0))=0$. From $P(x,0)$ we get $f(f(x))+f(0)=xf(0)+f(x)$. There are two cases. Case 1: $f(0)=0$. Then $f(f(x))=f(x)$. If $f(a)=f(b)\neq 0$, comparing $P(a,b)$ and $P(b,a)$ gives \[ af(b)+f(a+b)=f(ab+f(a))+f(b)=f(ab+f(b))+f(a)=bf(a)+f(a+b) \]so $a=b$. Thus $f$ is injective in $\mathbb{R}\setminus\{0\}$ so $f(x)\in\{0,x\}$. If $f(1)=0$, we have \begin{align*} P(x,1)&\implies f(x+f(x))=f(x+1)\\ P(1,x)&\implies f(x)+f(x)=f(x)+f(x+1) \end{align*}so $f(x+f(x))=f(x)$. Thus $f(x)=0$ for all $x$. Now assume $f(1)=1$. Assume for the sake of contradiction there exists $a\neq 0$ such that $f(a)=0$. Then $P(a,1)$ yields $1=a+f(a+1)$. Since $f(a+1)\in\{0,a+1\}$, it follows that $a=0$ or $a=1$, a contradiction. Thus $f(x)=x$ for all $x$. Case 2: $f(0)\neq 0$. Then \[ x=\frac{f(f(x))+f(0)-f(x)}{f(0)} \]so $f$ is injective. $P(x,f(0))$ gives $f(xf(0)+f(x))=f(x+f(0))$ so $f(x)=x(1-f(0))+f(0)$. Since $f(f(0))=0$, it follows that $f(0)(1-f(0))+f(0)=0$ so $f(0)=0$ or $f(0)=2$. The conclusion follows. $\square$