Find all $c\in \mathbb{R}$ such that there exists a function $f:\mathbb{R}\to \mathbb{R}$ satisfying $$(f(x)+1)(f(y)+1)=f(x+y)+f(xy+c)$$for all $x,y\in \mathbb{R}$. Proposed by Kaan Bilge
Problem
Source: Turkey Olympic Revenge 2023 P1
Tags: functional equation, function, function in R, olympic revenge, algebra
gghx
08.03.2023 12:25
Super nice problem!
We claim the only solution is $c=1$, in which $f(x)\equiv x$ works, and $c=-1$, in which $f(x)\equiv x^2$ works. Suppose now that $c\ne \pm1$, we show no $f$ satisfies the FE.
Let $P(x,y)$ be the assertion that $(f(x)+1)(f(y)+1)=f(x+y)+f(xy+c)$.
Note that $P(x,0)$ gives $(f(x)+1)(f(0)+1)=f(x)+f(c)$. If $f(0)\ne0$, then $f$ is constant, which is obviously not a solution. Hence, $f(0)=0$. This also means $f(c)=1$ (hence $c\ne0$). Compare $P(1,-1)$ and $P(c,-1)$. The former gives $(f(1)+1)(f(-1)+1)=f(c-1)$ while the latter gives $2(f(-1)+1)=f(c-1)$. Hence, either $f(1)+1=2$ or $f(-1)+1=0$.
Case 1: $f(1)+1=2\implies f(1)=1$.
Compare $P(x,1)$ and $P(x,c)$. The LHS are the same, hence $f(x+1)+f(x+c)=f(x+c)+f(xc+c)\implies f(x+1)=f(c(x+1))\implies f(x)=f(cx)$. Now $P(cx,cy)$ and using $f(x)=f(cx)$ gives $f(xy+c)=f(c^2xy+c)=f(cxy+1)$, or $f(x+c)=f(cx+1)$. But from above, $f(cx+1)=f(x+\frac{1}{c})$, hence $f(x+c)=f(x+\frac{1}{c})$. Let $c-\frac{1}{c}=a$ (note $a\ne0$, then $f(x)=f(x+a)$. $P(x+a,y)$ gives $f((x+a)y+c)=f(xy+c)$, which means $f$ is constant, contradiction.
Case 2: $f(-1)+1=0\implies f(-1)=-1$.
$P(1,-1): f(c-1)=0$. Now pick $a,b$ such that $a+b=c-1$ and $ab=-1$. We are guaranteed that neither $a$ nor $b$ are $-1$, else $c=1$. $P(a,b): (f(a)+1)(f(b)+1)=0$, so there exists $f(k)=-1$ where $k\ne -1$. Hence, $P(x,k): f(x+k)+f(xk+c)=0\implies f(x)=-f(xk+c-k^2)$. Furthermore, $P(x+1,-1): f(x)=-f(-x+c-1)$. Thus, $$f(xk+c-k^2)=-f(x)=f(-x+c-1)=-f((-x+c-1)k+c-k^2)=f(-[(-x+c-1)k+c-k^2]+c-1).$$Simplifying, $f(xk+c-k^2)=f(xk-ck+k+k^2-1)$. This means $f(x)=f(x+t)$ where $t=c-k^2-(-ck+k+k^2-1)=(k+1)(c-2k+1)$. By a similar logic to above, $a=0$, hence $(k+1)(c-2k+1)=0$. Note $k\ne -1$, so $k=\frac{c+1}{2}$. WLOG $a=\frac{c+1}{2}$, then $b=\frac{c-3}{2}$, so $\frac{c+1}{2}\frac{c-3}{2}=-1\implies (c-1)^2=0$, contradiction.
Hence, only $c=1,-1$ work.
omeravci372742
08.03.2023 23:10
Let us show that $c=\pm 1$. When we have $f(x)=x$ and $c=1$ or $f(x)=x^2$ and $c=-1$ equation holds. Now let us show that $c$ can not take any different value. If $f(0)\neq 0$ when we put $y=0$ we get $$f(x)f(0)=f(c)-f(0)-1$$Thus $f(x)$ is constant but then we put it into the equation we get $f(x)^2+1=0$ which is not possible. Thus $f(0)=0$. Then put $y=0$ then $f(c)=1$. So clearly $c\neq 0$. Put $x=1,y=-1$ then we get $$(f(1)+1)(f(-1)+1)=f(0)+f(c-1)=f(c-1)$$Put $x=c,y=-1$ then we get $$(f(c)+1)(f(-1)+1)=f(c-1)+f(0)=f(c-1)$$Then we either have $f(1)=f(c)$ thus $f(1)=1$ or $f(-1)=-1$. Let us start with the first case. \begin{enumerate}[(i)] \item We have $f(1)=1$. Put $y=1$ then $$2f(x)+2=f(x+1)+f(x+c)$$put $y=c$ then since $f(c)=1$ we get $$2f(x)+2=f(x+c)+f(cx+c)$$so we have $$f(x+1)=f(cx+c)$$then we can simply say $f(x)=f(cx)$. Then let us put $x=cx, y=cy$ then $$(f(cx)+1)(f(cy)+1)=f(cx+cy)+f(c^2xy+c)$$Clearly left hand side is equal to $(f(x)+1)(f(y)+1)$ thus it is also equal to $f(x+y)+f(xy+c)$. On the right hand side we have $f(cx+cy)=f(x+y)$ therefore we ultimately get $$f(c^2xy+c)=f(xy+c)$$then we have $$f(cxy+1)=f(c^2xy+c)=f(xy+c)=f(cxy+c^2)$$and by right choice of $x,y$ we can see that $f$ is periodic with $c^2-1$ so $f(x)=f(x+c^2-1)$ for all $x$. Then put $y=c^2-1$ so $f(y)=f(0)=1$ and we get $$f(x)+1=f(x+c^2-1)=f((c^2-1)x+c)=f(x)+f((c^2-1)x+c)$$Therefore $f((c^2-1)x+c)=1$ for all $x$. But then if $c\neq \pm1$ we get $f(x)$ constant but this is not possible as we mentioned before so we get $c=\pm 1 $ from this case. \item We have $f(-1)=-1$. Put $y=-1$ then $$0=(f(x)+1)(f(-1)+1)=f(x-1)+f(-x+c)$$Thus we have $f(c-x)=-f(x-1)$. Then put $y=1$ then $$(f(x)+1)(f(1)+1)=f(x+1)+f(x+c)=f(x+1)-f(-x-1)$$Then put $x=x-1$ so we get $$(f(x-1)+1)(f(1)+1)=f(x)-f(-x)$$Then put $x=-x$ so we get $$(f(-x-1)+1)(f(1)+1)=f(-x)-f(x)$$Add them together to get $$(f(x-1)+f(-x-1)+2)(f(1)+1)=0$$Then we either have $f(1)=-1$ or $f(x-1)+f(-x-1)=-2$. Let us analyze the two cases. \begin{enumerate}[(a)] \item Let $f(1)=-1$. Put $y=1$ then we have $$0=(f(x)+1)(f(1)+1)=f(x+1)+f(x+c)$$And put $x=x-1$ to get $f(x)=-f(x+c-1)$ then we have $f(x)=-f(x+c-1)=f(x+2c-2)$ thus $f$ is $2c-2$ periodic. If we have $c\neq 1$ then put $y=y+2c-2$ then we get $$(f(x)+1)(f(y+2c-2)+1)=f(x+y+2c-2)+f(xy+x(2c-2)+c)$$From periodicity we get left hand side is equal to $(f(x)+1)(f(y)+1)$ and on the right side we have $f(x+y)=f(x+y+2c-2)$ thus we get $$f(xy+c)=f(xy+c+x(2c-2))$$Then by properly choosing $x,y$ we can show that $f(x)$ is constant which is not possible. \item Then by putting $x=x-1$ we have $$f(x)+f(-x-2)=-2$$and we also had $$f(x)+f(-x+c-1)=0$$Then we have $$f(-x+c-1)=f(-x-2)+2$$and by putting $-x-2=x$ we get $$f(x+c+1)=f(x)+2$$Then let us put $y=1$ then we get $$(f(x)+1)(f(1)+1)=f(x+1)+f(x+c)$$then also put $x=x+c+1$ and we get $$((f(x+c+1)+1)(f(1)+1)=f(x+c+2)+f(x+2c+1))$$For right hand side we have $$f(x+c+2)+f(x+2c+1)) = f(x+1)+f(x+c)+4$$and for the left hand side we have $$((f(x+c+1)+1)(f(1)+1)=(f(x)+3)(f(1)+1)$$Then from the change of the left hand side and right hand side we have $2f(1)+2=4$ thus $f(1)=1$ and we are back to first case and we have already shown when $f(1)=1$ we have $c=\pm1$ then we are done. \end{enumerate} \end{enumerate}
zaidova
08.01.2024 22:05
We have two cases. f(x)=x or f(x)=x². But there will be a contrast like f(x)²+1=0. And f(1)=1. Then for P( -c; 1); (f(-c)+1)(f(1)+1)=f(-c+1)+f(0) For f(1)=1; f(0)=0 (f(-c)+1)*2=f(-c+1) 2*f(-c)+2=f(-c+1) And it only pays for c=1 And also we had a function before. f(x)=x² So c can be equal to +1; -1