Question

How many non-perfect square numbers lie between \[{n^2}\] and \[{\left( {n + 1} \right)^2}\]?

Answer 1

Hint: We will use that both \[{n^2}\] and \[{\left( {n + 1} \right)^2}\] are perfect square and they are consecutive perfect squares. Then we will have all the numbers between them are non-perfect square by finding the difference between \[{n^2}\] and \[{\left( {n + 1} \right)^2}\]. Then we will subtract the obtained equation by 1 to find the non-perfect square numbers.

Complete step by step answer:

We are given that the numbers are \[{n^2}\] and \[{\left( {n + 1} \right)^2}\].
We know that both \[{n^2}\] and \[{\left( {n + 1} \right)^2}\] are perfect square and they are consecutive perfect squares.
So, we will have all the numbers between them are non-perfect square.
Finding the difference between \[{n^2}\] and \[{\left( {n + 1} \right)^2}\], we get
\[ \Rightarrow {\left( {n + 1} \right)^2} - {n^2}\]
Using the trigonometric value, \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] in the above equation, we get
\[
   \Rightarrow {n^2} + 2 \cdot n \cdot 1 + {1^2} - {n^2} \\
   \Rightarrow {n^2} + 2n + 1 - {n^2} \\
   \Rightarrow 2n + 1 \\
 \]
Now we will subtract the above equation by 1 to find the non-perfect square numbers, we get
\[
   \Rightarrow 2n + 1 - 1 \\
   \Rightarrow 2n \\
 \]
Hence, there are \[2n\] non-perfect square numbers.

Note: We have subtract the result answer with 1 as we do not want to include the endpoints for non-perfect square numbers between \[{n^2}\] and \[{\left( {n + 1} \right)^2}\]. We need to know that a perfect square is an integer that is the square of an integer and a non-perfect square is not a square of any number.