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I have a problem and I think I know how to solve it so here it is: Determine $F(x)$ if, for all real $x$ and $y$, $F(x)F(y)-F(xy) = x+y$.

I tried a couple of cases and these were my results:

When $x=0$ and $y=0$ then $F(0) = 1$. When $y=0$ and $x$ is anything real, then $F(x) = 1 + x$ (or 1 + y if we did with $x=0$). So it seems like $F(x) = x+1$ works but I don't know if that is enough. Any hints or help would be greatly appreciated!

InsigMath
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    I would say slightly incomplete because taking $x=y=0$ also gives you the possibility $F(0)=0$. Can you investigate this possibility? – David Feb 13 '14 at 00:04
  • Well if we allow $F(0)=0$ then when one of the variables then we have a contradiction. So for instance let $x=n$ where $x$ is not 0 then $F(x)F(0)-F(0)=n$ or $n=0$. Which can't happen. – InsigMath Feb 13 '14 at 00:15
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    Fine, I think you now have a complete answer. – David Feb 13 '14 at 00:26

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As David notices, the $x=y=0$ case can also be satisfied by $F(0)=0$, and you need to show that this is not in fact possible.

Except for this gap, what you have is enough to prove that $F(x)=x+1$ is the only possible function that can satisfy the equation.

However you also need to show that $F(x)=x+1$ is in fact a solution, by substituting it into the equation and noticing that you get an identity after simplifying.

hmakholm left over Monica
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