Question

Prove that if n is a perfect square, then n + 2 is not a perfect square.

Short answer

\(n + 2\)is not a perfect square when it is given that \(n\) is a perfect square.

Step-by-step solution

Introduction

The purpose is to show that \(n + 2\) is not a perfect square, when it is given that \(n\)is a perfect square.

It is better to show by using contradiction.

Assume that, \(n + 2\)is a perfect square when it is given that \(n\) is a perfect square.

As \(n\)a perfect square, then there exists a positive integer \(a\)such that, \(n = {a^2}\).

As \(n + 2\)is a perfect square, then there exists a positive integer \(b\)such that \(n + 2 = {b^2}\).

Subtract \(n\)from \(n + 2\).

\(\left( {n + 2} \right) - \left( n \right) = 2\)

\(\left( {{b^2}} \right) - \left( {{a^2}} \right) = 2\)

\(\left( {b + a} \right)\left( {b - a} \right) = 2\)

Therefore, \(b + a = 2\)and \(b - a = 1\).

Adding the above obtained equations

To get the value of\(b\), add the two equations together

\(\left( {b + a} \right) + \left( {b - a} \right) = 3\)

\(2b = 3\)

\(b = \frac{3}{2}\) is not an integer.

This is contradiction, as \(b\)is an integer and contradicts the assumption that \(n + 2\) is a perfect square.

Therefore, \(n + 2\)is not a perfect square when it is given that \(n\)is a perfect square.