Integral Pairs Satisfying the Equation (xy times 15^{2018} 1): A Comprehensive Analysis

Introduction

When dealing with Diophantine equations, the problem of finding integral pairs that satisfy a given equation is of significant importance. This article explores the solution methods for the equation xy times 15^{2018} 1. We aim to simplify the equation, express it in terms of various mathematical transformations, and ultimately find the number of integral pairs (x, y).

Simplification of the Equation

First, let's denote N 15^{2018}. The given equation can be rewritten as:

xy times N 1

This simplifies to:

xy times 15^{2018} 1

We introduce a new variable s x times y and another p xy1, thus the equation becomes:

ps - 1 N

Transformation and Quadratic Equation

Let's rearrange this to express p in terms of s:

p frac{N}{s - 1}

Using the relation y s - x, we substitute back into the expression for p to get:

p x(s - x) - x sx - x^2

This yields a quadratic equation in terms of x:

x^2 - sx p 0

To ensure integer solutions for x, the discriminant of this quadratic must be a perfect square. The discriminant Delta is given by:

Delta s^2 - 4p

Substituting p frac{N}{s - 1} into the discriminant formula:

Delta s^2 - 4 times frac{N}{s - 1}

This equation can be rewritten as:

s^2 - k^2 4 times frac{N}{s - 1}

Analysis of (N 15^{2018})

Now let's analyze (N 15^{2018}). Given that (15 3 times 5), the prime factorization of (15^{2018}) consists only of the prime factors 3 and 5, each raised to the power of 2018. Therefore, the number of divisors dN of (15^{2018}) can be calculated using the formula for the number of divisors given a prime factorization:

N 3^{2018} times 5^{2018}

The total number of divisors is ((2018 1) times (2018 1) 2019 times 2019).

Determining Integral Solutions

For each divisor d of (15^{2018}), we set:

s d

p frac{15^{2018}}{d - 1}

Then we calculate the discriminant (Delta s^2 - 4p). If (Delta) is a perfect square, then the quadratic equation will have integer solutions.

Each valid (s) yields two solutions ((x, y)) and ((y, x)), unless (x y). Therefore, the total number of integral pairs ((x, y)) is twice the number of valid (s), plus the identity case if any.

Given 2019 divisors, we find pairs ((x, y)) and ((y, x)) for each valid (s), resulting in:

2 times 2019 4038

Thus, the total number of integral pairs is 4038.