solving diophantine equations

1. the factoring method

Solve in positive integers $1/x+1/y=1/pq$ , where $p$ and $q$ are prime.
Solution: $1/x<1/pq$ , so $x>pq$ .

$\begin{align*} \fr{x}+\fr{y}&=\fr{pq}\\ pq(y+x)&=xy \\ xy-pqy-pqx&=0 \\ (x-pq)(y-pq)&=p^2q^2. \end{align*}$

Considering the positive divisors of $p^2q^2$ we obtain:

$\begin{cases} x-pq=1\\ y -pq=p^2q^2 \\ \end{cases} \quad \begin{cases} x-pq=p\\ y -pq=pq^2 \\ \end{cases} \quad \begin{cases} x-pq=q\\ y -pq=p^2q \\ \end{cases} \quad \begin{cases} x-pq=p^2\\ y -pq=q^2 \end{cases}$

yielding the solutions (writing $r=pq$ ):
$(1+r,r(1+r))$ , $(p(1+q),r(1+q))$ , $(q(1+p),r(1+p))$ , $(p(p+q),q(p+q))$ ,
and their equivalents with $x$ and $y$ swapped. And the special case:

$\begin{cases} x-pq=p^2\\ y -pq=q^2 \end{cases}$

with the solution $(2r,2r)$ , i.e. 9 solutions in all.

$\fr{x}+\fr{y}=\fr{n},$

where $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ , has $(2\alpha_1+1)\cdots (2\alpha_k+1)$ solutions in positive integers.

The equation is equivalent to $\displaystyle (x-n)(y-n)=n^2,$ and $n^2=p_1^{2\alpha_1}\cdots p_k^{2\alpha_k}$ has $(2\alpha_1+1) \cdots (2\alpha_k+1)$ positive divisors.

(to be continued)

further reading
Andreescu, Andrica, Cucurezeanu – An Introduction to Diophantine Equations