Find Integer Solutions Challenge

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SUMMARY

The challenge involves finding all integer pairs \((p, q)\) that satisfy the equation \(1 + 1996p + 1998q = pq\). By rearranging the equation to \((p - 1998)(q - 1996) = 1997^2\), where \(1997\) is identified as a prime number, the solutions are derived. The complete solution set includes the pairs \((1, 1997^2)\), \((1997, 1997)\), \((1997^2, 1)\), \((-1, -1997^2)\), \((-1997, -1997)\), and \((-1997^2, -1)\).

PREREQUISITES
  • Understanding of integer equations and factorization
  • Familiarity with prime numbers, specifically the properties of prime factorization
  • Basic algebraic manipulation skills
  • Knowledge of mathematical notation and expressions
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  • Study properties of prime numbers and their applications in number theory
  • Learn about Diophantine equations and their solution methods
  • Investigate the implications of rearranging equations in algebraic contexts
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Mathematicians, students studying number theory, and anyone interested in solving integer equations and exploring the properties of prime numbers.

anemone
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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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