<s>[INST] How many integer solutions to a diophantine equation Starting with the equation:
$\frac{1}{a}+\frac{1}{b}=\frac{p}{10^n}$,
I reached the equation:
$10^{n-log(p)} = \frac{ab}{a+b}$.
Now given the positive integer $n$, for what integer values of $p$ would the value of:
$10^{n-log(p)}$,
be rational?
Also, given positive integers $n$ and $p$, how would we find positive integer solutions to $a$ and $b$ that satisfy the second equation, where:
$a ≤ b$,
And is it possible to determine, given $n$ and $p$, how many $a$ and $b$ solutions exist?
 [/INST] As @Qiaochu Yuan noted, 
$10^{n-\log p}=\dfrac{10^n}{p}$ is always a rational number.


Suppose that a positive integer $n$ and an integr $p$ are given, 
and let $a$ and $b$ to be integers 
satisfying the above relation: 
$$ 
\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{p}{10^n} 
\Longleftrightarrow 
\\ 
pab = 10^na + 10^nb 
\Longleftrightarrow 
\\ 
p^2ab = 10^npa + 10^npb 
\Longleftrightarrow 
\\ 
p^2ab - 10^npa - 10^npb = 0 
\Longleftrightarrow 
\\ 
p^2ab - 10^npa - 10^npb + 10^{2n} = 10^{2n} 
\Longleftrightarrow 
\\ 
(pa - 10^n) (pb - 10^n) = 10^{2n} 
. 
$$

notice that both of the factors 
$(pa - 10^n)$ and $(pb - 10^n)$ 
are (positive or negative) divisor of $10^{2n}$, 
so we can conclude that 
there is a (positive or negative) divisor $d$ of $10^{2n}$; 
such that: 
$$ 
(pa - 10^n)=d 
\ \ \ \ \ \ 
\text{and} 
\ \ \ \ \ \ 
(pb - 10^n)=\dfrac{10^{2n}}{d} 
\ \ \ \ \ \ 
\Longleftrightarrow 
$$ 
$$ 
a = \dfrac {d+10^n} {p} 
\ \ \ \ \ \ 
\text{and} 
\ \ \ \ \ \ 
b  = \dfrac {   \dfrac{10^{2n}}{d}   +10^n} {p} 
\ \ \ \ \ \ 
. 
$$
</s>