Step 1: Easy see that $f\equiv 0$ is a root. Now we will see at $f\not\equiv 0$. $f(xy+f(x))=xf(y)+f(x)(*)$ In (*) choose $x=y=-1$ then there exist $a$ such that $f(a)=0$, we have $a=0$. In fact, if $a\not = 0$ then choose $x=a,y=1$ have $f(1)=0$. Choose $y=1$ in (*) we have $f(x+f(x))=f(x)$ therefore $x+f(x)=x$ if $f(x)\not =0$. Contradiction! Note: If $f(m)=f(n)\not =0$ then $nf(m)+f(n)=mf(n)+f(m)$, so $m=n$. And we now have that $f(x)=0$ iff $x=0$ and $f$ is an injective function. P/S: olorin, where are you?

i'm not very good at functional equations so please correct me if i'm wrong. denote $f(0)=c$. $x=0$ gives $f(c)=c$. also, $y=0$ gives $f(f(x))=cx+f(x)\quad (1)$. substitute $x=c$ into $(1)$ to get $f(f(c))=c^{2}+c$. but we already have $f(f(c))=f(c)=c$. so $c^{2}+c=c\iff c=0$. so $(1)$ implies $f(f(x))=f(x)$. as N.T.TUAN proved that $f$ is injective, we have $f(x)=x$.

Oh, You are true! I think so

can you use fixed points to solve this? :

I don't think that method works in this problem!

this is my solution im not sur: if we exgeng x and y we take f(xy+f(y))=yf(x)+f(y) (1) x=0 and y=0 f0f(0)=f(0) (2) lets x and y sush that f(x)=f(y) we have f(xy+f(x))=xf(x)+f(x)* and (1) lmplay f(xy+f(x))=yf(x)+f(x)** * and ** implay x=y or f(x)=0 x=y implay that f is injective so from (2) we have f(0)=0 if we take y=0 in the initial equation we find f0f(x)=xf(0)+f(x)=f(x) f0f(x)=f(x) and f is injective so f(x)=x to sum up f(x)=x or f(x)=0

It is my solution!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! f(x*y+f(x))=x*f(y)+f(x) Denote f(0)=c; if y=0--> f(f(x))=x*c+f(x); (1) if y=f(y)-->f(f(xy+f(x)))=f(x*f(y)+f(x))=x*f(f(y))+f(x) ,but from (1) f(f(xy+f(x)))=xyc+f(x)*c+f(xy+f(x))=xyc+f(x)*c+x*f(y)+f(x); That's why xyc+f(x)*c+x*f(y)+f(x)=x*f(f(y))+f(x)=xyc+x*f(y)+f(x) --> f(x)*c=0 either f(x)=0 for x is from R or c=f(0)=0 -->f(f(x))=f(x); (*) Now we will prove that f(a)=f(b)--> a=b; f(ab+f(a))=a*f(b)+f(a)=f(ab+f(b))=b*f(a)+f(b); So f(a)*(a+1)=f(a)*(b+1)--> a=b from (*) f(f(x))=f(x)-->f(x)=x; So either f(x)=x or f(x)=0;

nayel wrote: $ f(f(x)) = f(x)$. as N.T.TUAN proved that $ f$ is injective, we have $ f(x) = x$. I don't understand this part. How does $ f$ injective and $ f(f(x)) = f(x)\forall x$ imply $ f(x) = x\forall x$? Wouldn't we need it to be surjective rather than injective? Anyway here's how I would finish the approach by N.T.TUAN: N.T.TUAN wrote: Easy see that $ f\equiv 0$ is a root. Now we will see at $ f\not\equiv 0$. $ f(xy + f(x)) = xf(y) + f(x)(*)$ Note: If $ f(m) = f(n)\not = 0$ then $ nf(m) + f(n) = mf(n) + f(m)$, so $ m = n$. In (*) choose $ x = y = - 1$ then there exist $ a$ such that $ f(a) = 0$, we have $ a = 0$. In fact, if $ a\not = 0$ then choose $ x = a,y = 1$ have $ f(1) = 0$. Choose $ y = 1$ in (*) we have $ f(x + f(x)) = f(x)$ therefore $ x + f(x) = x$ if $ f(x)\not = 0$. Contradiction! And we now have that $ f(x) = 0$ iff $ x = 0$ and $ f$ is an injective function. I shall only need that $ f(x) = 0$ iff $ x = 0$ Suppose $ x\neq 0$, $ y: = 1 - \frac {f(x)}{x}$. Then $ xy + f(x) = x\implies f(xy + f(x)) = f(x)\implies \text{(from (*))}\implies xf(y) = 0$ $ \implies f(y) = 0\implies y = 0\implies 1 - \frac {f(x)}{x} = 0\implies f(x) = x\forall x\neq 0$

That $ f$ is injective is equivalent to $ f(a)=f(b)\iff a=b\,\forall a,b\in\mathbb R$. Hence $ f(f(x))=f(x)$ implies $ f(x)=x$ (here $ a: =f(x)$ and $ b: =x$).

Thanks nayel for the explanation. I understand it now.

Isin't the easiest way of solving a function equation, to solve the equation itself. I mean we know that for every x : (f(f(f(f.....(k))))) = f(k) = k. Then we would have f(xy) + f(f(x)) = f(xy) + f(x) which implies to: xy + x =xy + x. So, knowing that for every x and y the equation is true, we also know that for every "n" it is true. The our function would be: f(n)=n. Am I right?

ggbg90 wrote: can you use fixed points to solve this? : The concept of fixed points could have been applied to this function if the question would have asked to prove that there are no polynomial solutions of this function of degree $\ge 2$.

N.T.TUAN wrote:

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$f(x+f(x))=f(x)$ therefore $x+f(x)=x$ if $f(x)\not =0$. Contradiction!

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Sorry to bring up an old topic,but I think my question is logical.In N.T.TUAN's approach for proving the injecitivity of $f$,I think he used the fact that $f$ is injective.

claim: $f$ is injective . proof: :suppose $f(a)=f(b)$ check $P(a,b) , P(b,a)$ we get : $ af(b)+f(a)=bf(a)+f(b) \implies a=b$ so let me back to main problem: $P(0,0) \implies f(f(0))=f(0) \implies f(0)=0$ check$P(x,0) \implies f(f(x))=f(x) \implies $ $ f(x)=x $

arzhang2001 wrote: claim: $f$ is injective . proof: :suppose $f(a)=f(b)$ check $P(a,b) , P(b,a)$ we get : $ af(b)+f(a)=bf(a)+f(b) \implies a=b$ so let me back to main problem: $P(0,0) \implies f(f(0))=f(0) \implies f(0)=0$ check$P(x,0) \implies f(f(x))=f(x) \implies $ $ f(x)=x $ But $f(x)=0$ is a solution

Ans: $f(x)=x$, $f(x)=0$. It is easy to see that these satisfy the equation, let $P(x,y)$ be the given assertion, we have \[P\left(x,\frac{x-f(x)}{x}\right)\implies xf\left(\frac{x-f(x)}{x}\right)=0 \ \ \ \forall x\ne 0\]\[P\left(\frac{x-f(x)}{x}, 0\right)\implies \left(1-\frac{x-f(x)}{x}\right)f(0)=0 \implies f(0)=0 \ \textrm{or} \ f(x)=0 \ \forall x\ne 0.\]If $f(x)=0 \ \forall x\ne 0$, then choose $x\ne 0,1$, \[P(x,0)\implies f(0)(1-x)=0\implies f(0)=0 \implies f(x)=0 \ \forall x\in \mathbb{R}.\]If $f(0)=0$, then we have $f\left(\frac{x-f(x)}{x}\right)=f(0)=0$, assume that we have $a,b\in \mathbb{R}$ such that $f(a)=f(b)=0$, $a\ne b$. Then, comparing $P(a,1)$ and $P(b,1)$ gives $af(1)=bf(1)=0\implies f(1)=0$, also $P(x,0)\implies f(f(x))=f(x)$. Now, $P(f(x),-1)\implies f(-1)f(x)=-f(x)$, we have already gotten $f(x)=0$, so $f(-1)=-1$ which further implies \[P(f(-1),x-1)\implies f(-x)=-f(x-1)-1.\]When $x=2$, $f(-2)=-1$, $P(-2,-2)\implies f(3)=1\implies f(3)=f(f(3))=f(1)=0$, a contradiction. $\blacksquare$

Let $P(x,y)$ the assertion of the given F.E. $P(0,0)$ $$f(f(0))=f(0)$$$P(f(0),0)$ $$f(0)^2=0 \implies f(0)=0$$$P(x,0)$ $$f(f(x))=f(x)$$$P(f(x),-1)$ $$f(x)(f(-1)+1)=0 \implies f(x)=0 \; \text{or} \; f(-1)=-1$$Now since $f(x)=0$ is a solution we will check what happens if $f(-1)=-1$. Let $a,b \ne 0$ be reals such that $f(a)=f(b)$ then by $P(a,b)-P(b,a)$ we have $a=b$ thus $f$ is injective. And with this we have thay since $P(x,0)$ gives $f(f(x))=f(x)$ we have by injectivity $f(x)=x$ thus we are done

Quote: by $P(a,b)-P(b,a)$ we have $a=b$ I got $af(a)=bf(a)$.

Let $P(x,y)$ be the assertion $f(xy+f(x))=xf(y)+f(x)$. Let $f(a)=f(b)$ for some $a,b$ such that $f(a)\ne0$. $P(a,b)-P(b,a)\Rightarrow af(a)=bf(a)$ Assume $f(0)\ne0$. $P(0,0)\Rightarrow f(0)=0$, contradiction. Hence $f(0)=0$. $P(x,0)\Rightarrow f(f(x))=f(x)$ If $f(x)\ne0$ then $f(x)=x$ by partial injectivity. Let $f(a)=0$ and $f(b)=b$ for some $a,b\ne0$: $P(b,a)\Rightarrow f(ab+b)=b\notin\{0,ab+b\}$, contradiction. Thus either $\boxed{f(x)=0}$ or $\boxed{f(x)=x}$.

$P(0,0): f(0)=0$. Suppose $f(a)=f(b)\ne0$. $P(a,b): f(ab+f(a))=af(a)+f(a)$. $P(b,a): f(ab+f(a))=bf(a)+f(a)$. Subtracting the two equations gives $a=b$. $P(x,0): f(f(x))=f(x)$, so $f(x)=x$ if $f(x)\ne0$. Thus, either $f(x)=0$ or $f(x)=x$. Suppose $f(a)=a$ and $f(b)=0$ and $a,b\ne0$. $P(a,b): f(ab+a)=a\implies ab+a=a\implies ab=0$, a contradiction. So the only solutions are $\boxed{f\equiv0}$ and $\boxed{f(x)=x\forall x\in\mathbb{R}}$, which both work.

Claim: $f$ is injective. Proof : Assume $f(a_1) = f(a_2)$. $P(a_1,a_2) : f(a_1a_2 + f(a_1)) = a_1f(a_2) + f(a_1)$. $P(a_2,a_1) : f(a_1a_2 + f(a_2)) = a_2f(a_1) + f(a_2)$. so we have $a_1 = a_2$. $P(0,0) : f(f(0)) = f(0) \implies f(0) = 0$. $P(x,0) : f(f(x)) = f(x) \implies f(x) = x$ or $f(x)= c$. for $x=0$ we have $f(0) = 0$ so $c = 0$. Claim: $f(x) = x$ or $f(x) = 0$. Proof : Assume $f(z_1) = z_1$ and $f(z_2) = 0$. $P(z_1,z_2) : f(z_1z_2 + z_1) = z_1$ for both cases $f(z_1z_2 + z_1) = 0$ and $f(z_1z_2 + z_1) = z_1z_2 + z_1$ we have contradiction cause $z_1$ and $z_2$ are not constant. Answer: $f(x) = x, f(x) = 0$.

1.assume $f$ nonconst then It is obvious see that $f$ injective $P(0,0)$ then $ff(0)=f(0)$ $P(f(0),0)$ then $f(0)=0$ $P(x,\frac{x-f(x)}{x})$ $f(\frac{x-f(x)} {x}) =0$ Then by injectivity $f(x) =x$ 2. Assume $f$ const then $f(x) =0$ And observe Pointwise trap.

$f(f(x))=xf(0)+f(x)$ Case 1:$f\equiv0$ Case 2:$f\not\equiv0$ Then we prove that $f$ is injective If $f(a)=f(b),a\ne b$ $bf(a)+f(b)=f(ab+f(b))=f(ab+f(a))=af(b)+f(a)$ So $b=a$ Then $f(f(0))=f(0)\implies f(0)=0$ Then $f(f(x))=f(x)\implies f(x)=x$

flower417477 wrote: $f(f(x))=xf(0)+f(x)$ Case 1:$f\equiv0$ Case 2:$f\not\equiv0$ Then we prove that $f$ is injective If $f(a)=f(b),a\ne b$ $bf(a)+f(b)=f(ab+f(b))=f(ab+f(a))=af(b)+f(a)$ So $b=a$ Then $f(f(0))=f(0)\implies f(0)=0$ Then $f(f(x))=f(x)\implies f(x)=x$ Not quite. You can only show injective for non-zero values of $f$. ($bf(a) = af(b) \implies f(b)= f(a)=0\text{ or }a=b$ Therefore in your last line you enter the point-wise trap $f(x)\in\{0,x\}$