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Find all $f:\mathbb{R}\rightarrow\mathbb{R}$ so that $f(xf(y)+x)=xy+f(x)$.

If you put $x=1$ it's easy to prove that f is injective. Now putting $y=0$ you can get that $f(0)=0$.

$y=\frac{-f(x)}{x}$ gives us $f(\frac{-f(x)}{x})=-1$

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  • $\begingroup$ I tried putting $y=x=1$ and got $f(f(1)+1)=1+f(1)$ from which I guessed that $f(x)=x$. This seems to fit the identity you give but doesn't seem like a strong enough proof. $\endgroup$ – Mufasa Jan 18 '15 at 14:10
  • $\begingroup$ It's a step in the right direction. Actually, if you put $f(x)=ax$ you get that $a=\pm 1$, so $f(x)=\pm x$ are for now the only slutions I have, but are not poven. $\endgroup$ – HeatTheIce Jan 18 '15 at 14:12
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Set $x = 1$ and $y = t - f(1)$,

then $f(1+f(t - f(1))) = t$, for all $t \in \mathbb{R}$, hence $f$ is a surjective function.

So, $\exists\, y_1 \in \mathbb{R}$, s.t., $f(y_1) = 0$

Then, $$f(x) = f(x+f(y_1)x) = xy_1+f(x) \implies xy_1 = 0, \forall x \in \mathbb{R} \implies y_1 = 0$$

and $\exists\, y_2 \in \mathbb{R}$, s.t., $f(y_2) = -1$

Then, $$0 = f(0) = f(x+xf(y_2)) = xy_2 + f(x)$$

i.e., $f(x) = -y_2x$ for $x \in \mathbb{R}$.

Plugging it back in the functional equation we may verify $y_2 = \pm 1$.

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Let f(1)=b. Putting now $x=1,y=-b$ we get that $f(-b)=-1$ Now we have that $f(\frac{-f(x)}{x})=-1=f(-b)$ so $f(x)=bx$. Putting that in the original eqation we get $b=\pm1$ and so is $f(x)=\pm x$ the solution.

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Substituting $y=1$, we get $f(xf(1)+x)=x+f(x)$.

If $f(1)=-1$, then $f(x)=-x$.

Assume $f(1)+1\ne 0$.

Let $p=f(1)+1$.

Assume $|p|<1$. The other case is similar with the substitution $x={1\over p}x$.

$f(p^nx)-{1\over p-1}p^nx=\dots=f(px)-{1\over p-1}px=f(x)-{1\over p-1}x$

Letting $n\to\infty$, we get $f(x)={1\over p-1}x$. I'm sure you can continue from here...

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