Find Integer Solutions Challenge

[anemone]

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

 

[Opalg]

Hint:
[sp]What is $(p-1)(q+1)$?[/sp]
Further hint:
[sp]$1997$ is a prime number.[/sp]

 

[kaliprasad]

I shall proceed differently from Opalg

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