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Find all functions $g$ from the Real numbers to itself, satisfying $g(x + y) + g(x)g(y) = g(xy) + g(x) + g(y) . .(*)$

This is a National Olympiad problem, however, my solution is quite different from the one provided by the author so I need you people to check my solution and tell me if its correct.

Solution:

setting $y = 0$ in $(*)$, we obtain

$g(x) + g(x)g(0) = 2g(0) + g(x)$, or $g(0)(g(x) - 2) = 0$.

So either $g(0) = 0$ or $g(x) = 2$, because$g(0) = 0$, is a valid solution, it follows that $g(x) = x$. $g(x) = 2$ are solutions as well.

Icosahedron
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Your solution is not correct. You have not ruled out the possibility that $g(0)=0$ but $g(x)\neq x$ for values of $x$ different from $0$. So it is not correct to conclude that $g(x)=x$ in the case $g(0)=0$.

Eric Wofsey
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  • $g(0) = 0$ satisfies the given equality, why would I want to rule it out? – Icosahedron Aug 17 '17 at 06:29
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    I never said anything about ruling out $g(0)=0$. You seem to be concluding that if $g(0)=0$ then $g$ must be the function $g(x)=x$, which is totally unjustified. – Eric Wofsey Aug 17 '17 at 06:41
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    No, you need to rule out the solutions $g(x)$ that satisfy both (1) $g(0) = 0$ and (2) $g(x)\neq x$ for some $x$ if you want to conclude that $g(x) = x$ is the only solution aside from $g(x) = 2$. – Michael L. Aug 17 '17 at 06:41