Solution: Let put it $y\rightarrow 0 \Longrightarrow f(f(x))=xf(0)+f(x)$ and we have $f(xy+f(x))=xf(y)+f(x)$ then put it $x\rightarrow f(x)$ thus $f(yf(x)+xf(0)+f(x))=f(x)f(y)+xf(0)+f(x)$ and put it $x,y\rightarrow 0$ and we know that $f(f(0))=0$ so that $f(0)=0$ and $f(f(x))=f(x)$ so substitute $y\rightarrow f(y)$ then $f(x)=x$

ehsan2004 wrote: Solution: Let put it $y\rightarrow 0 \Longrightarrow f(f(x))=xf(0)+f(x)$ and we have $f(xy+f(x))=xf(y)+f(x)$ then put it $x\rightarrow f(x)$ thus $f(yf(x)+xf(0)+f(x))=f(x)f(y)+xf(0)+f(x)$ and put it $x,y\rightarrow 0$ and we know that $f(f(0))=0$ so that $f(0)=0$ and $f(f(x))=f(x)$ so substitute $y\rightarrow f(y)$ then $f(x)=x$ The solution assumes that f is onto. I think that this is WAY too much to assume. ( Otherwise the step x --> f(x) cannot be justified.) Indeed, there are into functions satisfying the equation eg: f(x) = 0 for all x. Can someone please prove the ontoness if we exclude the identically vanishing function ( I think that f(x) = x or f(x) = 0 are the only solutions) or give a different solution? I have really tried hard but in vain...

If f(x) is linear --> f(x) = kx+b left = f(xy + f(x)) = f(xy + kx +b) = (y+k)x^2 + bx + b right = f(x) + xf(y) = kx + b + ky^2 + bx so, (y+k) x^2 = kx + ky^2 Assume y --> 0 so kx^2 = kx so that k = 1 OR 0 If n = deg(f(x)) is greater than one; deg(f(xy+f(x))) = 2n. for the left, deg(f(x) + xf(y)) = n +1 so for the right part, everything above degree n+1 should be null. Am I right? Then y = 0 again ... Frank Morgan

To post 4. Yes, I ave also seen that the first solution has some gasps. Really , f must be onto of << one to one>> , which is noway to find. Now , as about your solution: Do we know that f is a ponynomial ?Otherwise we can not assume << degree of function >>. Any way , I 'll try again to find something. Babis

Stergiu why you said that "f must be onto of << one to one>> , which is noway to find."

Kalimdor wrote: If f(x) is linear --> f(x) = kx+b left = f(xy + f(x)) = f(xy + kx +b) = (y+k)x^2 + bx + b right = f(x) + xf(y) = kx + b + ky^2 + bx so, (y+k) x^2 = kx + ky^2 Assume y --> 0 so kx^2 = kx so that k = 1 OR 0 If n = deg(f(x)) is greater than one; deg(f(xy+f(x))) = 2n. for the left, deg(f(x) + xf(y)) = n +1 so for the right part, everything above degree n+1 should be null. Am I right? Then y = 0 again ... Frank Morgan I don't think you are right because this ain't a polynomial so you can't suppose that it has a degree. Anyway I might be wrong so please correct me if so

ehsan2004 wrote: Stergiu why you said that "f must be onto of << one to one>> , which is noway to find." I think he means that your solution was not correct at last part:$f(x)f(y)+f(x)$ can't run over set of reals numbers!!

This problem is really easy but nice!! Do as ehsan2004, we have :$f(f(x))=f(x)$ I rewrite the initial relation:$f(xy+f(x))=xf(y)+f(x)$ Take $y$->$f(y)$ we obtain:$f(xf(y)+f(x))=xf(y)+f(x)$ And :$f(f(xy+f(x)))=f(xf(y)+f(x))$ ->$xy+f(x)=xf(y)+f(x)$.Therefore:$f(y)=y$ for all $y$ and we are done!

keira_khtn wrote: And :$f(f(xy+f(x)))=f(xf(y)+f(x))$ ->$xy+f(x)=xf(y)+f(x)$.Therefore:$f(y)=y$ for all $y$ and we are done! Can you explain why $xy+f(x)=xf(y)+f(x)$?,I am not sure about it ,also $f(x)=0$ solution too.

I think I've got a complete solution. \[ f(xy+f(x))=xf(y)+f(x) \] First we consider the case when $f(1)\not=0$. In this case, let $y_x=\frac{x-f(x)}{x}$ for every $x\not=0$. Then $xy_x+f(x)=x$ and substituting $y=y_x$ implies $xf(y_x)=0$ for every $x\not=0$. Hence $f(y_x)=0$. For every $x\not=0$, substituting $x=y_x, y=1$ in the original equation we get $0=f(y_x)=y_xf(1)$. Then $f(1)\not=0$ implies $y_x=0$. Hence $x-f(x)=xy_x=0$ and $f(x)=x$ for every $x\not=0$. It is then easy to check that $f(0)=0$ and thus $f(x)=x$ for every $x$. Now consider the second case that $f(1)=0$. We claim that $f=0$. Suppose not, say $f(t)\not=0$. Consider $y_t=\frac{t-f(t)}{t}\not=1$. Substituting $x=y_t, y=0$ we have $f(0)=y_tf(0)$ and thus $f(0)=0$. Now, substituting $y=0$ and $y=1$ we may get $f(f(x))=f(x)$ and $f(x+f(x))=f(x)$ Let $s=f(t)\not=0$. Then $f(s)=f(f(t))=f(t)=s$, $f(2s)=f(s+f(s))=f(s)=s$, and $f(3s)=f(2s+f(2s))=f(2s)=s$. Now substituting $x=s, y=2$ in the original equation we get $f(2s+f(s))=sf(2)+f(s)$, which implies $f(2)=0$ Finally, substituting $x=2, y=s$ we get $f(2s)=2f(s)$, which leads to $s=0$, contradiction!

it's not hard to prove that f is injective. Let f(x)=f(y) then xy+f(x)=xy+f(y) hence f(xy+f(x))=f(xy+f(y)) hence we get xf(y)+f(x)=yf(x)+f(y) hence we get x=y

https://artofproblemsolving.com/community/c6h152023p855649