# Find functions $f,g$ such that $f(x+y)=g(x·y)$ for all $x,y$? [closed]

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.

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

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## 1 Answer

$$f(x) = f(x + 0) = g(x \cdot 0) = g(0)$$ 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.

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