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