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Find all functions $f: \mathbb R\rightarrow \mathbb R$, at the same time satisfying the following two conditions:

a) $f (x + yf (x)) = f (x) f (y)$

b) the function $f$ can be represented in the form $f (x) = (\varphi (x)) ^ 2, x \in \mathbb R,$ where the function $f$ has a finite derivative at $x = 0.$ (not infinite)

I have no clue how to start. Any kind of help will be appreciated.

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  • $\begingroup$ What do you mean by a 'finite derivative'? Non zero, not infinite? $\endgroup$
    – copper.hat
    Apr 22 '16 at 16:13
  • $\begingroup$ @ copper.hat: not infinite $\endgroup$
    – Roman83
    Apr 22 '16 at 16:15
  • $\begingroup$ $f(0)=f(0)^2$ $f(0)=0$ $\endgroup$
    – kmitov
    Apr 22 '16 at 16:16
  • $\begingroup$ Is $f(x)=0$ and $f(x)=1$ a solution? Doesn't solve the problem, though. $\endgroup$
    – S.C.B.
    Apr 22 '16 at 16:18
  • $\begingroup$ @kmitov also $f(0)=1$ is possible $\endgroup$
    – gt6989b
    Apr 22 '16 at 16:21
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Here is a possible approach.

Plug in $y=0$ to get $$ f(x) = f(x) f(0), $$ so either $f(x) \equiv 0$ or $f(0)=1$.

Now note that $$ f(x+yf(x)) = f(y+xf(y)) $$ and assuming $f$ is 1-to-1, we have $$ x + yf(x) = y + xf(y)\\ x(1-f(y)) = y(1-f(x))\\ \frac{x}{1-f(x)} = \frac{y}{1-f(y)} $$ for arbitrary $x,y$, and that means both LHS and RHS and constant, say $c$.Then you have $$ c = \frac{x}{1-f(x)} \\ f(x) = 1 - x/c $$

UPDATE If $f$ is not 1-1, $f(x) \equiv 1$ is a solution, but not sure if there are others...

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  • $\begingroup$ Why $$f(x+yf(x)) = f(y+xf(y)) \Rightarrow x + yf(x) = y + xf(y)$$? $\endgroup$
    – Roman83
    Apr 22 '16 at 16:38
  • $\begingroup$ @Roman83 if you assume $f$ is one-to-one, then $f(a) = f(b) \implies a = b$. Notice the result satisfies this property as well as that $f(0)=1$... $\endgroup$
    – gt6989b
    Apr 22 '16 at 16:41

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