MHB Find Integer Solutions Challenge

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The challenge is to find integer pairs (p, q) satisfying the equation 1 + 1996p + 1998q = pq. By rearranging the equation, it can be expressed as (p-1998)(q-1996) = 1997^2. The solutions derived from this equation are (1, 1997^2), (1997, 1997), (1997^2, 1), along with their negative counterparts (-1, -1997^2), (-1997, -1997), and (-1997^2, -1). The prime nature of 1997 is crucial in determining the solution set. Thus, the integer solutions are fully characterized by these pairs.
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Find all pairs $(p, q)$ of integers such that $1+1996p+1998q=pq$.
 
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Hint:
[sp]What is $(p-1)(q+1)$?[/sp]
Further hint:
[sp]$1997$ is a prime number.[/sp]
 
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
 

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