TITLE: Functions satisfying $f(a+b) =\frac{f(a)+f(b)}{f(a)*f(b)}$
QUESTION [5 upvotes]: i was looking for a function that satisfies $f(a+b) =\frac{f(a)+f(b)}{f(a)*f(b)}$ for all $a,b \in \Bbb{N}$. I have never seen such a problem before and i would like some kind of help to get me started.

REPLY [2 votes]: Assuming $0\in \mathbb{N}$ we get $f(0)=\pm \sqrt 2$ as noted in comments.  

Let $f(0)=\sqrt 2$

Then $f(1)=\frac{f(1)+f(0)}{f(1)\cdot f(0)}=\frac{f(1)+\sqrt 2}{\sqrt 2 \cdot f(1)}$ leads to the quadratic equation $\sqrt 2 \cdot [f(1)]^2-f(1)-\sqrt 2 =0$ wich has solutions $f(1)=\sqrt 2\;$ and $\;f(1)=-\frac{\sqrt 2}{2}.$ 

$f(1)=\sqrt 2\;$ leads necessarily to $$\;f(a)=\sqrt 2\quad\text{for}\quad a\in \mathbb{N},$$ because the sequence defined as $a_1=\sqrt 2,\; a_{n+1}=\frac{a_n+\sqrt 2}{\sqrt 2\cdot a_n},\; n\in \mathbb{N}\;$ is constant.
Consider $f(1)=-\frac{\sqrt 2}{2}.$    We get $$f(2)=\frac{2}{f(1)}=-2\sqrt2,\quad f(3)=\frac{f(2)+f(1)}{f(2)f(1)}=-\frac{5\sqrt 2}{4},$$ but then we would obtain two different values of $f(4):$ $$f(4)=\frac{2}{f(2)}=-\frac{\sqrt2}{2}\quad \text{or}\quad f(4)=\frac{f(3)+f(1)}{f(3)f(1)}=-\frac{7\sqrt2}{5}$$ 

Therefore $f(1)=\sqrt2$ and the function is constant.

Let $f(0)=-\sqrt 2$
Similarly to the above, $f(1)$ can take two values.    

$f(1)=-\sqrt2$ determines a constant solution $f(n)=-\sqrt2,$ 
the other is $f(1)=\frac{\sqrt2}{2}$ and leads to a disambiguation.


Conclusion
The only solutions defined in $0$ are the constant functions $f(n)=\sqrt2\;$ and $g(n)=-\sqrt 2, n=0,1,\;\dots$