Aug 1, 2013 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,967 Find all pairs $(p, q)$ of integers such that $1+1996p+1998q=pq$.
Aug 1, 2013 Moderator #2 Opalg MHB Oldtimer Staff member Feb 7, 2012 2,820 Hint: Spoiler What is $(p-1)(q+1)$? Further hint: Spoiler $1997$ is a prime number.
Aug 2, 2013 #3 kaliprasad Well-known member Mar 31, 2013 1,358 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
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