Functional Equation in the rationals: $f(xy)=f(x)f(y)-f(x+y)+1$

Asked Jul 22 '20 at 04:40

Active Dec 18 '20 at 13:31

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Find all the possible functions $f:\mathbb{Q}\to\mathbb{Q}$ such that $$f(1)=2$$ and $$f(xy)=f(x)f(y)-f(x+y)+1\text.$$

I managed to find the function for the set of natural number, by putting $x=1$ and $y=n$. From it, I have that $$f(n+1)=f(n)+1\text,$$ which means that $f(n)$ is an arithmetic progression of natural numbers.

I am stuck with generalizing this into the set of rational number, any hit will be appreciated.

edited Dec 18 '20 at 13:31

Mohsen Shahriari

6,718

asked Jul 22 '20 at 04:40

The game

One more: https://math.stackexchange.com/questions/96325/finding-the-function – VIVID Jul 22 '20 at 05:05
how can I find if my question is duplicate – The game Jul 22 '20 at 05:06
You can check here: https://approach0.xyz/search/?q=%24f(xy)%3Df(x)f(y)-f(x%2By)%2B1%24&p=1 – VIVID Jul 22 '20 at 05:07
@VIVID that's a duplicate, a triplicate, and a quadruplicate! – Gerry Myerson Jul 22 '20 at 06:30

Functional Equation in the rationals: $f(xy)=f(x)f(y)-f(x+y)+1$

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