Question

Prove that there are 100 consecutive positive integers that are not perfect squares. Is your proof constructive or non constructive?

Short answer

There are 100 consecutive positive integral values that are not perfect square.

And, the proof is constructive.

Step-by-step solution

Introduction

Consider the statement that there are 100 consecutive positive integral values that are not perfect squares.

The purpose is to prove the statement and check whether it is constructive or not.

Proof using constructive method

Consider the perfect squares \({{\bf{n}}^{\bf{2}}}\)and \({\left( {{\bf{n}} + 1} \right)^{\bf{2}}}\)where \({\bf{n}}\) is positive integer.

It is important to note that there are \(2n\)positive integers strictly between them because of the following:

\(\begin{aligned}{}{\left( {n + 1} \right)^{\bf{2}}} - {n^2} = {n^2} + 2n + 1 - {n^2}\\ = 2n + 1\end{aligned}\)

Take \(n = 50\).

\(\begin{aligned}{}{n^2} = {50^2}\\ = 2500\end{aligned}\)

And,\(\begin{aligned}{c}{\left( {n + 1} \right)^2} = {51^2}\\ = 2601\end{aligned}\)

Note that 2501, 2502,….2600 meet the criteria that have no perfect square.

That is there are 100 consecutive positive integral values that are not perfect square.

And, the proof is constructive.