chaotic_iak wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that \[f(x+y) + f(x)f(y) = f(xy) + (y+1)f(x) + (x+1)f(y)\] for all $x,y \in \mathbb{R}$. Steps of my solution: (Is it correct? ) Let $P(x,y)$ be the assertion $f(x+y) + f(x)f(y) = f(xy) + (y+1)f(x) + (x+1)f(y)$ $P(0,0) \Rightarrow f(0)=0 $ or $f(0)=2$ a) $f(0)=2$ : $P(x,0) \Rightarrow f(x)=x+2$ but it dosen't work!! b)$f(0)=0$ Define $f(1)=c$ so $P(1,1) \Rightarrow f(2)=5c-c^2$ $p(2,2,) \Rightarrow c=0$ or $c=2$ or $c=3$ or$c=5$ i)$f(1)=0 \Rightarrow $ $P(x,1) \Rightarrow f(x+1)=3f(x) \Rightarrow f(n)=0 \forall n \in \mathbb{Z}$ $P(x,-1) \Rightarrow f(-x)= \frac 13 f(x)$ (*) $P(x,n) \Rightarrow f(nx)= (3^n-n-1)f(x) \Rightarrow f(x)=0 \forall x \in \mathbb{Q}$ $P(x,2) \Rightarrow f(2x)=6f(x)$ (**) $P(x,-x) , P(x,x) ,* ,**\Rightarrow f(x)=0 $ ii)$f(1)=2$ $P(x,1) \Rightarrow f(x+1)=f(x)+2x+2 \Rightarrow f(n)=n(n+1) \forall n \in \mathbb{Z}$ $P(x,-1) \Rightarrow f(-x)= f(x-1) =2x+f(x)$ (*) $P(x,2) \Rightarrow f(2x)=4f(x)-2x$ (**) $P(x,-x) , P(x,x) ,* ,**\Rightarrow f(x)=x(x+1) $ iii)$f(1)=3$ $P(x-1,1) \Rightarrow f(x)=3x$ iv) $f(1)=5$ $P(1,1) \Rightarrow f(2)=0$ $P(x-1,1) \Rightarrow f(x)=5x-2f(x-1) \Rightarrow f(-1)=0$ $P(x,-1) \Rightarrow f(-x)=\frac{5x-f(x)}{2}$ (*) $P(x,2) \Rightarrow f(2x)=f(x)-5x$ (**) $P(x,-x) , P(x,x) ,* , ** \Rightarrow f(x)(3x-1) = 5x(x+1) $ but $x=\frac 13 $ is contradictiuon! So answers are $f(x)=0, f(x)=x(x+1), f(x)=3x$

chaotic_iak wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that \[f(x+y) + f(x)f(y) = f(xy) + (y+1)f(x) + (x+1)f(y)\] for all $x,y \in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)$ Let $f(0)=a$ Let $f(1)=b$ $P(0,0)$ $\implies$ $a(a-2)=0$ If $a=2$, then $P(x,0)$ $\implies$ $f(x)=x+2$ which is not a solution. So $a=0$ $P(2,2)$ $\implies$ $f(2)(f(2)-6)=0$ and so $f(2)\in\{0,6\}$ $P(1,1)$ $\implies$ $f(2)=5b-b^2$ and so $b\in\{0,2,3,5\}$ 1) $b=0$ $P(x,1)$ $\implies$ $f(x+1)=3f(x)$ Comparing then $P(x,y)$ and $P(x,y+1)$, we get $f(xy+x)=3f(xy)+(2y+1)f(x)$ which may be written $f(u+v)=3f(u)+(2\frac uv+1)f(v)$ $\forall u,v\ne 0$ Swiching $u,v$, we get $3f(u)+(2\frac uv+1)f(v)=3f(v)+(2\frac vu +1)f(u)$ and so $\frac{f(v)}v=-\frac{f(u)}u$ $\forall u\ne v$ and $u,v\ne 0$ Hence $f(x)=0$ $\forall x$ which indeed is a solution. 2) $b=2$ $P(x,1)$ $\implies$ $f(x+1)=f(x)+2x+2$ Comparing then $P(x,y)$ and $P(x,y+1)$, we get $f(xy+x)=f(xy)+f(x)(2y+1)-2xy$ which may be written $f(u+v)=f(u)+f(v)(2\frac uv+1)-2u$ $\forall u,v\ne 0$ Swiching $u,v$, we get $f(u)+f(v)(2\frac uv+1)-2u$ $=f(v)+f(u)(2\frac vu+1)-2v$ and so $\frac{f(v)}{v^2}-\frac 1v=\frac{f(u)}{u^2}-\frac 1u$ $\forall u,v\ne 0$ And so $f(x)=x^2+x$ which indeed is a solution 3) $b=3$ $P(x-1,1)$ $\implies$ $f(x)=3x$ which indeed is a solution 4) $b=5$ $P(-1,1)$ $\implies$ $f(-1)=0$ $P(x,1)$ $\implies$ $f(x+1)=-2f(x)+5x+5$ Comparing then $P(x,y)$ and $P(x,y+1)$, we get $f(xy+x)=-2f(xy)+(2y+1)f(x)-5xy$ Setting there $x=y=-1$, we get $f(0)=-2f(1)-5$ which is wrong. So not a solution. Hence the answers : $f(x)=0$ $\forall x$ $f(x)=3x$ $\forall x$ $f(x)=x^2+x$ $\forall x$ edit : too late !

pco wrote: edit : too late ! Thanks PCO...Now i am sure about my solution!

goldeneagle wrote: pco wrote: edit : too late ! Thanks PCO...Now i am sure about my solution! How do you get from the rationals to the real numbers?