Help with proving the surjectivity of a function satisfying $ f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $ I have the following functional equation for injective $ f : \mathbb R ^ * \to \mathbb R ^ * $
$$ x + y \ne 0 \implies f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $$
Found $ f ( 2 ) = \frac 1 2 $ and I am trying to find $ f ( 1 ) $, so I let
$$ x + y = x y \implies y = \frac x { x - 1 } $$
$$ f \left( \frac x { x - 1 } \right) = 1 - f ( x ) $$
then declared $ g : \mathbb R ^ * \setminus \{ 1 \} \to \mathbb R ^ * \setminus \{ 1 \} $
$$ g(x) = \frac{x}{x-1} $$
and was able to prove $ g $'s bijectivity. So, if I can prove $ f $ is surjetive (or alternatively that it is multiplicative or that, at least $ \exists x \, f ( x ) = 1 $) it follows that $ f ( 1 ) = 1 $, $ \forall q \in \mathbb Q \, f ( q ) = q ^ { - 1 } $ and from this point it should be easy to finish the problem.
Am I going the wrong path? Is there an easier way to deal with this problem? Can someone give me hints?
Update:
I was able to make a system of equations with $ f ( - 1 ) $, $ f  ( 1 ) $, $ f ( 2 ) $ and $ f ( - 2 ) $, and found $ f ( - 1 ) = - 1 $, $ f ( - 2 ) = \frac { - 1 } 2 $, $ f ( 1 ) = 1$ and $f \left( \frac 3 2 \right) = \frac 2 3 $. Not as elegant as the method I wanted to use but it works.
Using the identity with $ f \circ g ( x ) $ got the values for $ f \left( \frac k { k - 1 } \right)$ plugging positive integers and $ f \left( \frac { k - 1 } k \right) $ with negative ones. Now I am doing some manipulation so I can get from this to $ \frac 1 k $ and from there it should be easy to induce for the rationals. I will post my calculations soon.
The problem is, having no info about continuity, how could we expand this to the reals? That's harder than I thought it would be. We do know, however, the function is decreasing for positive rational numbers. This may be useful.
Second Update:
I got it, the secret is to look at $ \frac 1 { f ( x ) } $, will probably answer this question myself next week.
 A: You can show that the only functions satisfying (I will assume that $ x , y \ne 0 $ everywhere I use them)
$$ x + y \ne 0 \Longrightarrow f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) \tag 0 \label 0 $$
where $ f : \mathbb R ^ * \to \mathbb R ^ * $, are $ f ( x ) = \frac 1 x $ and $ f ( x ) = \frac 1 2 $. It's obvious that these two are solutions. To show that these are the only solutions, we define $ g : \mathbb R ^ * \to \mathbb R ^ * $ such that $ g( x ) = \frac 1 { f ( x ) } $. Then \eqref{0} yields
$$ x + y \ne 0 \Longrightarrow \frac { g ( x ) + g ( y ) } { g ( x + y ) } = \frac { g ( x ) g ( y ) } { g ( x y ) } \tag 1 \label 1 $$
It's worth mentioning that \eqref{1} tells us that if $ x + y \ne 0 $ then $ g ( x ) + g ( y ) \ne 0 $. Now, letting $ x = y = 2 $ in \eqref{1} we get $ g ( 2 ) = 2 $. Putting $ x = 2 $ and $ y = - 1 $ gives us $ \frac { g ( - 1 ) + 2 } { g ( 1 ) } = \frac { 2 g ( - 1 ) } { g ( - 2 ) } $, and letting $ x = y = -1 $ we'll have $ \frac { 2 g ( - 1 ) } { g ( - 2 ) } = \frac { g ( - 1 ) ^ 2 } { g ( 1 ) } $. Combining these two equations, we get $ g ( - 1 ) = - 1 $ or $ g ( - 1 ) = 2 $.
Let's suppose that $ g ( - 1 ) = 2 $. Then by the above equations $ g ( - 2 ) = g ( 1 ) $. Now, letting $ x = 1 $ and $ y = -3 $ in \eqref{1} we have $ g ( 1 ) + g ( - 3 ) = g ( 1 ) ^ 2 $, while $ x = - 1 $ and $ y = - 2 $ gives us $ 2 + g ( 1 ) = g ( 1 ) g ( - 3 ) $. Combining these two and rearranging the terms, we get $ \big( g ( 1 ) - 2 \big) \big( g ( 1 ) ^ 2 + g ( 1 ) + 1 \big) = 0 $, and hence $ g ( 1 ) = 2 $. Therefore, if $ x \ne - 1 $ then letting $ y = 1 $ in \eqref{1} we have $ g ( x + 1 ) = 1 + \frac { g ( x ) } 2 $, or equivalently, we have $ g ( x - 1 ) = 2 \big( g ( x ) - 1 \big) $, for $ x \ne 1 $. Now, if $ x \ne 1 $ then letting $ y = - 1 $ in \eqref{1} we'll have $ g ( - x ) = \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } $. Substituting $ - x $ for $ x $ in the last equation, and using the equation itself, one can get
$$ g ( x ) = \frac { 4 \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } \left( \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } - 1 \right) } { \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } + 2 } $$
for $ x \ne -1 , 1 $. Straightforward algebraic manipulation yields $ 3 g ( x ) \big( g ( x ) - 2 \big) \big( 10 g ( x ) ^ 2 - 5 g ( x ) - 2 \big) = 0 $.
$ g ( x ) $ can't be equal to $ \frac { 5 \pm \sqrt {105} } { 20 } $ since in that case the value of $ g ( x + 1 ) = 1 + \frac { g ( x ) } 2 $ or $ g ( x - 1 ) = 2 \big( g ( x ) - 1 \big) $ would be unacceptable. Thus we have $ g ( x ) = 2 $ for $ x \ne -1 , 1 $, which easily shows that $ f ( x ) = \frac 1 2 $ for every $ x $.
Now, suppose $ g ( - 1 ) = - 1 $. This shows that $ g ( - 2 ) = - 2 g ( 1 ) $. Using this fact, if we let $ x = 1 $ and $ y = - 3 $ in \eqref{1} we get $ g ( 1 ) + g ( - 3 ) = - 2 g ( 1 ) ^ 2 $, and for $ x = - 1 $ and $ y = - 2 $ we have $ - 1 - 2 g ( 1 ) = g ( 1 ) g ( - 3 ) $. Combining these equations we have $ \big( 2 g ( 1 ) + 1 \big) \big( g ( 1 ) + 1 \big) \big( g ( 1 ) - 1 \big) = 0 $. If $ g ( 1 ) = - \frac 1 2 $ then by the above equation $ g ( - 3 ) = 0 $ which can't happen. We can also rule out the case $ g ( 1 ) = - 1 $ by considering different values of $ x $ and $ y $ in \eqref{1}, which I omit. Thus we have $ g ( 1 ) = 1 $.
Now, if $ x \ne - 1 $, letting $ y = 1 $ in \eqref{1} we have $ g ( x + 1 ) = g ( x ) + 1 $, or equivalently $ g ( x - 1 ) = g ( x ) - 1 $ for $ x \ne 1 $.
Thus letting $ x = - 1 $ in \eqref{1}, we get $ g ( - x ) = - g ( x ) $ for $ x \ne 1 $, which can easily be generalized to include every $ x $, and thus $ g $ is an odd function. Now, using a simple induction, we can get $ g ( x + n ) = g ( x ) + n $, for every nonnegative integer $ n $ and $ x \ne - n , - n + 1 , \dots , - 1 $. Letting $ y = n $ in \eqref{1}, we get $ g ( n x ) = n g ( x ) $ for $ x \ne - n , - n + 1 , \dots , - 1 $, since by a simple induction we have $ g ( n ) = n $. This together with $ g $ being odd, shows that in fact for every $ x $ and every nonzero integer $ n $, $ g ( n x ) = n g ( x ) $. Now letting $ y = x $ in \eqref{1}, we get $ g \big( x ^ 2 \big) = g ( x ) ^ 2 $. This shows that if $ x > 0 $ then $ g ( x ) > 0 $, and since $ g $ is odd, the converse is also true. Now, for every two integers $ m $ and $ n $, if $ m \ne 0 $ and $ m x + n \ne 0 $, then we have $ m x + n > 0 $ iff $ g ( m x + n ) > 0 $. Thus $ m x + n > 0 $ iff $ m g ( x ) + n > 0 $. This means that $ x $ and $ g ( x ) $ must be equal, since otherwise we can find a rational number between them, which will give a counterexample to what we've just proven. Thus in this case we have $ f ( x ) = \frac 1 x $ for every $ x $.
