Question

Prove that there are no solutions in integers \(x\) and \(y\) to the equation\(2{x^2} + 5{y^2} = 14\).

Short answer

This equation \(2{x^2} + 5{y^2} = 14\) has no solutions.

Step-by-step solution

Introduction

Here in this question use exhaustive proof this includes entertaining all possibilities.

Proof using exhaustive method

Consider the equation as given

\(2{x^2} + 5{y^2} = 14\).

Take different ranges for x and y values and analyse the value of the equation and find out whether there is a solution or not,as follows;

\(2{x^2} + 5{y^2} = 14\)

When;

\(\begin{aligned}{l}\left| x \right| \ge 3;\,2{x^2} > 14\\\left| y \right| \ge 2;5{y^2} > 14\end{aligned}\)

Hence,\(\left| x \right|\)can’t take values more than 3 and \(\left| y \right|\)can’t take values more than 2.

So, xcan take values\(\left( {{\bf{ - 2, - 1,0,1,2}}} \right)\)and y can take values\(\left( {{\bf{ - 1,0,1}}} \right)\).

So possible values of\(2{x^2}\)are 0, 2, 8 and possible values of\(5{y^2}\)are 0, 5. Hence the maximum addition which is possible can be\(8 + 5 = 13\).

Hence this equation has no solutions.

Therefore, the result stated in the question has been proved.