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My approach towards this question was that first I putted x=0 and then y=0 which yields f(x)=f(y)and f(x)=g(0), and again for x=1 & y=1 it gives g (x)=f(1+x), and so on. My query is that what should the compact & rigorous form of proof or solution looks like because I found myself stuck between results and not comprehending results very well. Thanks.

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closed as unclear what you're asking by user21820, José Carlos Santos, YiFan, Alexander Gruber Apr 29 at 23:44

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$$f(x) = f(x + 0) = g(x \cdot 0) = g(0)$$ for all $x$. Thus $f(x)$ is constant and always equal to $g(0)$. Moreover, $$g(0) = f(1 + y) = g(1 \cdot y) = g(y)$$ for all $y$. Thus $g$ is also constant and equal to $g(0)$. Thus $f, g$ are the same constant function. All constant functions $f = g = c$ will satisfy the relation. This is all you need.

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    $\begingroup$ Yeah thanks man m looking for this kind of preciseness. $\endgroup$ – BORN TO LEARN Apr 29 at 4:05
  • $\begingroup$ Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x). $\endgroup$ – BORN TO LEARN Apr 29 at 4:23
  • $\begingroup$ Now I think your proof also needs editing? $\endgroup$ – BORN TO LEARN Apr 29 at 4:25
  • $\begingroup$ @BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y). $\endgroup$ – BenB Apr 29 at 4:26
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    $\begingroup$ @BORNTOLEARN because f is identically equal to g(0) as I show in the previous part. $\endgroup$ – BenB Apr 29 at 4:44

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