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This question already has an answer here:

I need help with this:

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

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marked as duplicate by Ross Millikan, rschwieb, Emily, Asaf Karagila, Stefan Hansen Feb 14 '13 at 6:20

This question was marked as an exact duplicate of an existing question.

I note that this is not quite a duplicate, as the variables are different. It may be that the techniques of the cited answer apply as well. Certainly $f(x)=g(x)=x$ is again a solution. – Ross Millikan Feb 13 '13 at 23:09

For the first, note that the right side is independent of $y$, so $f(x)$ is constant. Then we have $f(0)=g(f(0)+x)$ with the left side independent of $x$. Then $g$ must be constant, but it must be injective. No solution.

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