A Prime Denominator

We can rearrange the equation:

$\dfrac{1}{2017}=\dfrac{2(a+b)}{ab} \Longrightarrow ab=2 \cdot 2017(a+b).\;$ Let then $p=ab$ and $s=a+b$ .

Since $p(x)=x^2-sx+px$ has two integer roots, precisely $a$ and $b$ , it must be $\Delta=q^2$ for a certain $q \in \mathbb{Z}$ , so that $s^2-4p=q^2$ . Also, we have $\displaystyle b=\frac{s+q}{2}, \; a=\frac{s-q}{2}$ . Recalling then from above that $p=2 \cdot 2017s$ , it follows:

$s^2-8 \cdot 2017s=q^2$

$(s-q-4 \cdot 2017)(s+q-4 \cdot 2017)= (4 \cdot 2017)^2$

$(a-2 \cdot 2017)(b-2 \cdot 2017) = (2 \cdot 2017)^2$

So, we narrowed our search to couples of integers $A,B : A \cdot B = 2^2 2017^2$ , where $A+2 \cdot 2017=a$ and $B+2 \cdot 2017=b.\;$ For such couples, $0<a\le b \Rightarrow -2 \cdot 2017<A \le B$ . However, if $A<0$ , then $-2 \cdot 2017<A \Rightarrow \frac{(2 \cdot 2017)^2}{A}<\frac{(2 \cdot 2017)^2}{-2 \cdot 2017}$ and so $B<-2 \cdot 2017$ , which cannot be accepted, as just stated. The same goes for $B$ . Thus, $A,B>0$ .

There are $3 \cdot 3 =9$ positive divisors of $2^2 \cdot 2017^2$ , and hence there are $4$ couples of distinct $A,B$ and $1$ couple with $A=B=2\cdot 2017$ .

So, $4+1=\boxed{5}$