Diophantine Equations – Can you solve this?-1

Learn Diophantine equations step by step: how to find integer solutions, with worked examples on linear equations, Pythagorean triples, and Pell’s equation. This article is part of Number Theory where Diophantine equations are discussed. Here we will be solving an excellent example of linear Diophantine equation of type pA+qB-AB=0. We will be solving for A+B=? We discuss the most important factoring trick in solving this type of Diophantine Equations. Here not only the topic of Diophantine equations is discussed but also inequalities, prime factorization, finding divisors of a number, factoring tricks are discussed at length while solving a typical problem here. This type of concepts will be discussed in upcoming lecture notes like this.

Diophantine Equations – Problem -1

Find a+b=? if { $\left\{(a,b)\in {\mathbb{N}}^2:(\frac{1}{a}+\frac{1}{b})=\frac{1}{2026}\right\}$}

when a,b<10000.

Solution:

To understand the above Diophantine equation problem properly is required for solving it. let us break this problem into solvable parts.

Common Mistake to avoid

The common mistake to avoid in understanding above problem is,

$(a,b)\in {\mathbb{N}}^2$means $a\in \mathbb{N},b\in \mathbb{N}$and not $a\in {\mathbb{N}}^2,b\in {\mathbb{N}}^2$

Problem simplification

The given problem is simplified to the form,

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2026},a,b<10000,a,b\in \mathbb{N}$

From above, 2026(a+b)=ab

$2026(a+b)=\mathrm{ab}\Rightarrow \mathrm{ab}-2026a-2026b=0$

Factoring Trick- Most important step?

Application of factoring trick at this stage to solve the given Diophantine equation is the most important step in this problem solution.

$\mathrm{ab}-2026a-2026b=0\Rightarrow \mathrm{ab}-2026a-2026b+{2026}^2={2026}^2$

From above it can be deduced that,

$(a-2026\mathrm{)(}b-2026)={2026}^2$

Real Solution

Let (a-2026)=x, then

$(a-2026)=x\Rightarrow x(b-2026)={2026}^2\Rightarrow (b-2026)=\frac{{2026}^2}{x}$

From above we get,

$a=x+2026,b=\frac{{2026}^2}{x}+2026$

Application of inequalities

Given that $a,b<10000,a,b\in \mathbb{N}$.

$a<10000\Rightarrow x+2026<10000\Rightarrow x<7974$

$b<10000\Rightarrow \frac{{2026}^2}{x}+2026<10000\Rightarrow \frac{{2026}^2}{x}<7974$

From above both inequalities , we get,

$\frac{{2026}^2}{7974}<x<7974$

From above deductions we get ,

$\frac{{2026}^2}{7974}<x<7974\Rightarrow 514.76<x<7974\Rightarrow 515<x<7974$

Application of Divisors concept

${2026}^2=2^2\times {1013}^2\Rightarrow \mathrm{Divisors}=(2+1(2+1)=9$

As above the number of divisors of ${2026}^2$ = 9.

The 9 divisors of ${2026}^2$ are as below,

1,2,4,1013,2026,4052,1026169,2052338,4104676

Here it is to be noted that x is one of the divisors of ${2026}^2$ ans also $515<x<7974$.

Hence to get the value of x, we have to find divisors between 515 and 7974.

Hence the values that can be used for x are 1013,2026 and 4052 such that $a,b<10000,a,b\in \mathbb{N}$.

If x=1013, then ,

a=x+2026=1013+2026=3039

b= $\frac{{2026}^2}{x}$ 2026 = 2×2026+2026= 6078.

Hence a+b=3039+6078= 9117

As above for x=1013, the values of a and b are found out. Similarly we we can get the values of pairs (a,b) for x=2026 and 4052 as (4052,4052) and (6078,3039). The values of (a+b) are 8104 and 9117.

Conclusion

The above critical problem is solved by using the concepts of following principles of Number Theory,

1. Diophantine Equation
2. Factoring Trick
3. Application of Inequalities
4. Finding the number of divisors
5. Finding real divisors
6. Prime factorization principles

A typical image of the solution is presented here as,

The video solution for above problem is available on you tube channel maths&science by iitian, accessible at link :

https://www.youtube.com/@mathsciencebyiitian

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Best of luck to everybody.