What's the solution of the functional equation?

I need help with this:

"Find functions $f$, $g : \mathbb{Z} \rightarrow \mathbb{Z}$, knowing that $g$ is injective and such that: $$f(g(x)+y) = g(f(x)+x), \mbox{ for all } x, y \in \mathbb{Z}.$$

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Did you maybe mean $g(f(x)+y)$, or something to that effect? Otherwise set $x=0$ to get $f(g(0)+y)=g(f(0))$, so that $f$ is constant. –  Alex R. Feb 5 '13 at 20:46

There must be an error on the question. As the second member doesn't depend on $y$ then $f$ must be a constant $c$.
$f(g(0) + y) = g(f(0)) = c, \mbox{ for all } y \in \mathbb{Z}$
Then the first member of the equation is a constant, $f(g(x) + y)=c$, so $g$ must also be constant, $c = g(c + x) \mbox{ for all } x \in \mathbb{Z}$.
Maybe you mean $f(g(x)+y) = g(f(y)+x), \mbox{ for all } x, y \in \mathbb{Z}$ ?