I think my solution is shorter than you : we first prove $ f(1) = 1$ and $ f$ is injective and $ f(xy) + f(x) + f(y) = f(x + y) + f(x)f(y)$ (as the way MellowMelon shows) now if $ x = 1$ then $ f(x + 1) = f(x) + 1$ . now $ x + 1 \longrightarrow x$ which gives : $ f(xy + y) + f(x + 1) + f(y) = f(x + y + 1) + f(x + 1)f(y)$ now $ f(x + 1) = f(x) + 1$ and $ f(x + y + 1) = f(x + y) + 1$ which gives : $ f(xy + y) + f(x) + f(y) = f(x + y) + f(x)f(y) + f(y)$ $ \longrightarrow f(xy + y) - f(y) = f(x + y) + f(x)f(y) - f(x) - f(y)$ but the $ RH$ is $ f(xy)$ so we have : $ f(xy + y) = f(y) + f(xy)$ but we can put $ xy = a$ and $ y = b$ for all reals $ a,b$ and then : $ f(a + b) = f(a) + f(b)$ and then by $ f(xy) + f(x) + f(y) = f(x + y) + f(x)f(y)$ u can get : $ 1)f(a + b) = f(a) + f(b)$ $ 2)f(ab) = f(a)f(b)$ so $ f(x) = x$ for all $ x$ or $ f(x) = 0$ but $ f(x) = 0$ is not an answer so we have $ f(x) = x$

http://www.mathlinks.ro/viewtopic.php?t=187785 http://www.mathlinks.ro/viewtopic.php?search_id=1361591936&t=178957

What are these links? The problem is different from these links. BaBaK Ghalebi wrote: http://www.mathlinks.ro/viewtopic.php?t=187785 http://www.mathlinks.ro/viewtopic.php?search_id=1361591936&t=178957

Yeah! but the problem can be reduced to that functional equation, so that's probably what he meant.

MellowMelon wrote: Put in $ x = 0$. We get $ f(f(y)) = y(1 - f(0)) + f(0)$ I don't think it is true(maybe a typo) but the solution is correct (it is not affected ).For x=0 you get $ f(0) + y + f(0) = f(f(y)) + yf(0)$ or $ f(f(y)) = y(1 - f(0)) + 2f(0)$

Here is my solution,check it please.Thanks a lot.May be there are some similar ideas in the above solutions. Quote: $ f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)$ (*) Setting $ x = 0$ gives $ f(f(x)) = (1 - a)x + 2a$,where $ a = f(0)$. If $ a = 1$: Then $ f(f(x)) = 2 \forall x \in R$ and $ f(0) = 1$. Since $ f(0) = 1,$,we have $ f(f(0)) = f(1)$=>$ f(1) = 2$=>$ f(f(1)) = f(2)$=>$ f(2) = 2$ So $ f(1) = f(2) = 2$. Set $ x = 1$ into (*),we get $ f(f(y) + 1) = 4 - y$,so if $ f$ is injective,contradiction.(because we have $ f(1) = f(2)$) So $ a$ is not equal to 1. $ f(f(x)) = (1 - a)x + 2a$ and $ f$ is injective. Since RHS of the above condition=$ R$,there exists $ x_0$ such that $ f(x_0) = 0$ Set $ y = x_0$ into (*),we get $ x_0f(x) = x_0 + a$. If $ x_0$ is not equal to $ 0$,then $ f(x)$ is a constant.No solution in this case. Hence $ x_0 = 0$,this gives us $ f(0) = 0$ and $ f(f(x)) = x$. Replacing $ x$ by $ f(x)$ into (*) and using the condition $ f(f(x)) = x$,we get: $ f(f(x)f(y)) + x + y = f(f(x) + f(y)) + xy$ Setting $ x = f(x),y = f(y)$ into the above condition,we get:

to continue.

A very good solution , MellowMelon , congratulation ! The functional equation $ f(x + y) + f(x)f(y) = f(x) + f(y) + f(xy)$ is a result that should be remebered Using the lemma : if $ f: R \ \to \ R ; f(xy) = f(x)f(y) ; f(x + y) = f(x) + f(y) \ \forall \ x ; y \ \in \ R$ then $ f(x) = 0 \forall x \ \in \ R$ or $ f(x) = x \forall x \ \in \ R$ We can solve an other nice problem that had appeared in M&Y Magazine : Find all function $ f: R \ \to \ R$ such that : $ f(x - y) + f(xy) = f(x) - f(y) + f(x)f(y) \ \forall \ x ; y \ \in \ R$ Try with that problem:wink:

I liked this problem very much! Find all functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$ such that $f(xf(y))+y+f(x) = f(x+f(y))+yf(x)$ Let $P(x,y)$ be the assertion

So, $f(x)=x\forall x\in\mathbb{R}$ is the only solution.

Omid Hatami wrote: Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$: \[ f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)\] We will prove that $f(0)=0$. Let $f(0)=a$ and $f(1)=b$. In $(1)$ replacing $x=y=0$ : \[f(a)=2a\] In $(1)$ replacing $x=0,y=1$ : \[f(b)=a+1\] In $(1)$ replacing $x=1,y=0$ : \[f(a+1)=f(a)+b\] But $a+1=f(b)$ and $f(a)=2a$, we have : \[f(f(b))=2a+b\;\;\;\;\;(*)\] In $(1)$ replacing $x=0,y=b$ : \[f(f(b))=2a+b-ab\;\;\;(**)\] From $(*)(**)$ we deduce : \[2a+b-ab=2a+b\]. Implies $a=0$ or $b=0$. If $b=0$, we have $a=f(0)=f(b)=a+1\;\;\;\;(!)$. Hence $a=0$ or $f(0)=0$. Replacing $x=0$ in $(1)$ : \[f(0)+y+f(0)=f(f(y))+yf(0)\] Implies $f(f(y))=y$. In $(1)$ replacing $y$ by $f(y)$ and remembering $f(f(y))=y$ : \[f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)\] This is a familiar problem. See http://julielltv.wordpress.com/2014/08/15/functional-equation-58/

Very happy I solved this.

Oops this solution was longer than it should have been

Omid Hatami wrote: Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$: \[ f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)\]

This problem is trivial! We claim that $f \equiv x \ \forall x \in \mathbb{R}$ is the only solution.Denote by $P(x,y)$ the given assertion.Then, $$\rightarrow P(0,y) \implies 2\cdot f(0) + y = f(f(y)) + y\cdot f(0).$$Now, if $f(0)=1$, then we easily get that $f$ is a constant function, which is not true. Thus, we conclude that $f(0)\neq 1$ and consequently, $f$ is a bijection. Now, by surjectivity of $f$, we know that there exists a real number $\alpha$ such that $f(\alpha)=0$. Then, using $P(x,\alpha)$, we get that $\alpha = 0$. Thus, $f(0)=0$. Hence, $f(f(y))=y \forall y \in \mathbb{R}$. Then, $$P(x,f(y)) \implies f(xy)+f(x)+f(y)=f(x+y)+f(x)\cdot f(y)$$Now, plug in $x=y=2$, we get that $f(2)=2$.Then, since $f(f(1)+1)=f(2)$, it follows that $f(1)=1$.Thus, $P(x,1)$ gives $$f(x+1)=f(x)+1$$Replace $x \rightarrow x+1$ in $P(x,y)$ and get: $f(xy+y)+f(x)+1+f(y)=f(x+y)+1+f(x)\cdot f(y) + f(y) \iff f(xy+y) = f(xy)+f(y)$ Now, this gives $f$ is both additive and multiplicative and hence the conclusion follows.

Omid Hatami wrote: Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$: \[ f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)\] Taking p(1,f(1)\f(1)-1),for $f(1)$ not equal to $1$,we see that $1=0$ ,which is a contradiction,so now we see that $f(1)=1$. $p(0,1)$ $f(0)+1=f(0)=f(f(1))+f(0)$ . . $f(0)=0$., and from $p(0,x)$ we get that $f(f(x))=x$-involutive ,also this gives us f-injective + surjective . $p(x,1)$ gives us $f(x+1)=f(x)+1=f(x)+f(1)$. $p(x,f(y))$,gives us $f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)$ $p(x,f(y+1))$,gives us $f(xy+x)+f(y+1)+f(x)=f(x+y+1)+f(y+1)f(x)$ $f(xy+x)+f(y)+1+f(x)=f(x+y)+1+f(y)f(x)+f(x)$ this follow using $p(x,1)$ $f(xy+x)+f(y)=f(x+y)+f(x)f(y)$- Now use $p(x,f(y))$ to finish this part $f(xy+x)+f(y)=f(xy)+f(y)+f(x)$ $f(xy+x)=f(xy)+f(x),--> f(m+n)=f(m)+f(n)$ Now geting back to $p(x,f(y))$ we have $f(xy)+f(y)+f(x)=f(x+y)+f(x)f(y)$ $f(xy)+f(x)+f(y)=f(x)+f(y)+f(x)f(y)$ $f(xy)=f(x)f(y)---->$ multiplicative And now from multiplicatve + additive ,we get that the desired result to be f(x)=x $\blacksquare$

Omid Hatami wrote: Find all functions $ f: \mathbb R\longrightarrow \mathbb R$ such that for each $ x,y\in\mathbb R$: \[ f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)\] Let $P(x,y)$ denote the above assertion. $$P(x,0): f(xf(0))+f(x)=f(x+f(0))$$$$P(0,x): f(f(x))=x(1-f(0))+2f(0)$$now if $f(0)=1$ then using $P(x,0)$ we have $f(n)=2^n \forall n \in \mathbb{N}$ which is clearly impossible by substituting in the parent equation. Thus, $f(0) \ne 1$ and hence $f$ is bijective by $P(0,x) \implies$ there exists $c$ such that $f(c)=0$ $$P(x,c): f(0)+c=cf(x)$$clearly $f$ is not a constant function, so $f(0)=0$, $P(0,x): f(f(x))=x$ $$P(x,f(y)): f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)$$call this new assertion $Q(x,y)$ $$Q(x,1): f(x+1)=(2-f(1))f(x)+f(1)$$if $f(1) \ne 1$ then by an easy induction $f(n)=\frac {f(1)(2-f(1))^n}{1-f(1)}-\frac {f(1)}{1-f(1)}$ which can easily be dispelled. Thus $f(1)=1$ and by an easy induction using $Q(x,1)$ we have $f(n)=n \forall n \in \mathbb{Z}$ also $f(x+n)=f(x)+n$ and so by using $Q(x,n)$ we have $f(nx)=nf(x) \forall n \in \mathbb{Z} \implies f(-x)=-f(x)$ $$Q(x,-y)+Q(x,y): 2f(x)=f(2x)=f(x+y)+f(x-y) \implies f(a+b)=f(a)+f(b) \quad \forall a,b \in \mathbb{R}$$also by $P(a,b): f(ab)=f(a)f(b)$. Thus $f$ is both multiplicative and additive, so $f(x)=x \quad \forall x \in \mathbb{R}$ $\blacksquare$

We easily have that $f$ is nonconstant. $\textbf{1. }f(1)=1$ If $f(1)\ne1$ then $P\left(\frac{f(1)}{f(1)-1},1\right)\Rightarrow1=0$, contradiction. $\blacksquare$ $\textbf{Case 1: }f(0)=1$ $P(0,x)\Rightarrow f(f(x))=2$ $P(2,f(x))\Rightarrow f(2)+f(x)=f(2)f(x)$, so $f(2)=1$ since $f$ is nonconstant, but this implies that $f(2)=0$, contradiction. $\blacksquare$ $\textbf{Case 2: }f(0)\ne1$ $P\left(\frac{f(0)}{f(0)-1},0\right)\Rightarrow f\left(\frac{f(0)}{f(0)-1}\right)=0$, so let $k=\frac{f(0)}{f(0)-1}$. $P(x,k)\Rightarrow f(0)+k=kf(x)$, we must have $k=0$, since $f$ is nonconstant. So $f(0)=0$, hence $P(0,x)\Rightarrow f(f(x))=x$. $\blacksquare$ $\textbf{2. }f(x+y)=f(x)+f(y)$ $P(x,f(y))\Rightarrow Q(x,y):f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)$ $Q(x,1)\Rightarrow f(x)+1=f(x+1)$ $Q\left(x,\frac yx+1\right)-Q\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)\forall x\ne0$, but since it obviously holds for $x=0$, we're done here. $\blacksquare$ Substituting this back in to $Q(x,y)$, we easily get $f(xy)=f(x)f(y)$, and the only multiplicative and additive function is the identity function, $\boxed{f(x)=x}$, which fits. $\square$

Let $P(x,y)$ denote $f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)$. If $f(0)=1$, then $P(0,0)$ gives $f(1)=2$ and $P(0,1)$ and $P(1,0)$ gives contradiction. Now $P(\tfrac{f(0)}{f(0)-1},0)$ gives $f(x_0)=0$ for some $x_0$. $P(x,x_0)$ gives $f(0)=0$, since constant $f$ fails. So $f(f(x))=x$ from $P(x,0)$. $P(2,f(2))$ gives $f(2)=2$ and then $P(1,f(1))$ gives $f(1)=1$. Now $P(x,1)$ gives $f(x+1)=f(x)+1$. Now comparing $P(x,f(y+1))$ and $P(x,f(y))$ gives $f$ is additive. Now $P(x,f(y))$ gives $f$ is multiplicative. So $f\equiv \text{Id}$, which fits.

\[ f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)\]Only such function is $f(x)=x$ which clearly works. Claim: $f$ is an involution. Proof. Note that $f$ is non-constant. Plugging $x=0$ gives $y(1-f(0))+2f(0)=f(f(y))$. Now write $f(x)$ instead of $x$ in order to get \[f(f(x)f(y))+x-xf(0)+2f(0)+y=f(f(x)+f(y))+xy(1-f(0))+y.2f(0)\]Thus, $f(f(x)f(y))-f(f(x)+f(y))-xy(1-f(0))+x+y+2f(0)=f(0)(x+2y)$ and since left hand side is symmetric, we observe $(x+2y)f(0)=(y+2x)f(0)$ hence $f(0)=0$ which implies $f(f(x))=x$.$\square$ Claim: $f(1)=1$. Proof. Since $f$ is an involution, it's injective. $x=y=1$ yields $f(f(2))=2=f(f(1))+1=f(1+f(1))$ hence $f(1)+1=f(2)$. We also have \[f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)\]Pick $x=y=1$ which implies $3f(1)=f(2)+f(1)^2=f(1)^2+f(1)+1$ or $(f(1)-1)^2=0$ or $f(1)=1$.$\square$ Now, $y=1$ gives $f(x+1)=f(x)+1$. Compare $x,y$ with $x+1,y$ to see \[f(xy+y)+f(x)+f(y)+1=f(x+y)+1+f(x)f(y)+f(y)\]Thus, $f(xy+y)=f(xy)+f(y)$ or $f(x+y)=f(x)+f(y)$ hence $f$ is additive. By utilising this, we can conclude that $f$ is also multiplicative since $f(xy)+f(x)+f(y)=f(x+y)+f(x)f(y)=f(x)+f(y)+f(x)f(y)$. The solution of this Cauchy function is $f(x)=cx$ and plugging back yields $f(x)=x$ as desired.$\blacksquare$

Ashutoshmaths wrote: I liked this problem very much! Find all functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$ such that $f(xf(y))+y+f(x) = f(x+f(y))+yf(x)$ Let $P(x,y)$ be the assertion

How did he concluded that the function is injective and non-constant?