How to Find Pairs of Positive Integers in an Equation and Analyze Systems of Equations

When working with equations involving positive integers, it's essential to approach the problem systematically. This article delves into finding pairs of positive integers in the given equation and analyzes the resulting system of equations. Whether you're a student, a mathematician, or just curious about solving complex equations, this guide will provide you with a valuable toolkit for tackling similar problems.

Introduction to the Problem

Consider the equation:

( (xy)(xy)^2 – 7xy -20 )

Observations and Analysis

Let's begin by making some observations:

The given equation can be rewritten as: ( (xy)(xy)^2 - 7xy -20 ) Since ( xy ) must be positive, ( x, y > 0 ). If ( x 2 ) and ( y ) is any odd number, then ( 3x^3 10 ), which is not possible in the set of natural numbers ( N ). Since ( xy ) must be even, both ( x ) and ( y ) must be even or one must be even and the other odd. In either case, ( xy ) will be even, making ( (xy)^2 - 7xy ) odd, which would violate the given equation. The product ( xy ) must be at least 4 since ( xy

From these observations, we can deduce that ( xy ) can either be 4 or 10.

Deducing Solutions

We can now analyze the two systems of equations that arise:

( xy 4, (xy)^2 - 7xy -5 ) ( xy 10, (xy)^2 - 7xy -2 )

It turns out that only the first system yields consistent solutions, which are:

Solutions: ( boxed{(1, 3), (3, 1)} )

Analysis of Extended Equations

If ( xy p q ), then:

( x^2 - y^2 p^2 - 2q )

( x^3 - y^3 p^3 - 3q 4pq - 5 )

Substituting, we get:

( p^3 - 7pq 20 0 )

When ( q 3 ), we have:

( p^3 - 21p 20 0 )

This gives us possible values of ( pq -5, 3, 13, 43 ).

To find exact solutions, we can use the following system of equations:

( x^3 - y^3 4x^2y - xy^2 - 5 )

Let ( a xy )

( a(x^2 - y^2) 4axy - 20 )

( xy(x - y)^2 4xy^2 - 20 )

Considering the factors of -20, we get:

( xy 1 ) ( xy 2 ) ( xy 4 ) ( xy 5 ) ( xy 10 ) ( xy 20 )

For ( xy 4 ) and ( x^2 - 5xy y^2 -5 ), this results in:

( 16 - 28 3 -28 7 -5 )

Thus, the solutions are ( boxed{(1, 3), (3, 1)} ).

By systematically testing each factor, we can confirm that the only integer solutions are ( (1, 3) ) and ( (3, 1) ).