Q13/079) Find all pairs (p,q) of integers such that 1+1996p+1998q=pq.

pq – 1996p – 1998q = 1

Or (p-1998)(q-1996) – 1998 * 1996 = 1

Or (p-1998)(q-1996) = 1998 * 1996 + 1 = 1997^2

We get all the solution set for (p-1998, q- 1996) to be ( 1, 1997^2), (1997,1997), ( 1997^2, 1)
(-1, - 1997^2), (-1997,- 1997), (- 1997^2, - 1) as 1997 is prime