Functions satisfying $f(a+b) =\frac{f(a)+f(b)}{f(a)*f(b)}$ 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.
 A: It is not always clear whether $\mathbb N$ includes $0$ or not. If we ignore zero, then one way of working is to set $f(1)=x$ whence:
$$f(2)=f(1+1)=\frac 2x$$
$$f(3)=f(2+1)=\frac {x^2+2}{2x}$$
$$f(4)=f(3+1)=\frac {3x^2+2}{x^3+2x}$$but also $$f(4)=f(2+2)=x$$
In fact we have $f(2a)=f(a+a)=\frac 2{f(a)}$ so that $f(a)f(2a)=2=f(2a)f(4a)$ so that always $f(4a)=f(a)$
Equating the expressions for $f(4)$ gives an equation which can be solved for possible values of $x$
A: Asume that $f$ is defined on ${\mathbb N}_{\geq1}$, and let $f(1):=c$ for a $c\in\dot{\mathbb C}$ to be determined. From
$$f(n+1)={1\over c}+{1\over f(n)}\qquad(n\geq1)$$
we compute
$$\bigl(f(n)\bigr)_{1\leq n\leq 5}=\left(c,{2\over c},{2+c^2\over 2c},{2+3c^2\over c(2+c^2)},{2+5c^2+c^4\over c(2+3c^2)}\right)\ .\tag{1}$$
Now we should have
$$f(4)=f(2+2)={2 f(2)\over \bigl(f(2)\bigr)^2}={2\over f(2)}=c\ .$$
Comparing this with the value of $f(4)$ in $(1)$ we obtain the equation
$$c={2+3c^2\over c(2+c^2)}$$
with the solutions $\pm i$ and $\pm\sqrt{2}$. Starting anew with
$$f(5)=f(2+3)=\ldots\ ,$$
and comparing with the $f(5)$ from the table $(1)$ leads again to $c=\pm\sqrt{2}$ and to four complex $c$-values $\ne\pm i$. It follows that necessarily $c\in\{\pm\sqrt{2}\}$. It is then easily checked that the function $f(n):=c$ for all $n\geq1$ satisfies the given functional equation.
A: 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 $g(n)=-\sqrt 2, n=0,1,\;\dots$
