Question

If a natural number m can be expressed as \[{{n}^{2}}\], where n is also a natural number, then m is called:
(a) m is a square number
(b) m is not a square number
(c) m is not a perfect number
(d) m is a perfect number

Answer 1

Hint: First of all, try to recollect what natural number, square number, and perfect numbers are. Now, assume a natural number n and find \[{{n}^{2}}=m\] and see to what category it belongs.

Complete step-by-step answer:
In this question, we are given a natural number m that can be expressed as \[{{n}^{2}}\] where n is also a natural number. Then we need to tell if m is a square number, perfect number or none of them. Before proceeding with the question, let us understand the same terminologies.
Natural number: Natural numbers are the numbers that start from 1, 2, 3, and go till infinity. In other words, we can also say that all positive integers are natural numbers. Example: 1, 2, 3, 5000, 546, etc.
Square Numbers: Square numbers are numbers that are obtained by squaring a natural number. In other words, we can also say that these are the product of a number multiplied by itself. Examples: \[{{2}^{2}}=4;{{5}^{2}}=25;{{10}^{2}}=100,etc.\]
Perfect Number: Perfect numbers are positive integers that are equal to the sum of their positive divisors excluding the number itself. For example: 6 has divisors 1, 2 and 3 excluding itself and 6 = 1 + 2 + 3 [Sum of its divisors]. So, 6 is a perfect number. We have some more perfect numbers like 28, 496, etc.
Now, let us consider our question. In this question, we are given that n is a natural number. So, let us assume, n = 4. We are given that m can be expressed as \[{{n}^{2}}\]. So, we get,
\[m={{n}^{2}}={{\left( 4 \right)}^{2}}\]
m = 16.
This is also a natural number.
Let us find the divisors of 16. We know that,
\[\begin{align}
  & 16=1\times 16 \\
 & 16=2\times 8 \\
 & 16=4\times 4 \\
\end{align}\]
So, we get the divisors of 16 as 1, 2, 4, 8, 16. Let us find the sum of its divisors excluding itself. So, we get,
Sum of the divisors excluding itself = 1 + 2 + 4 + 8 = 15.
We know that \[15\ne 16\], that is the sum of its divisors excluding itself is not equal to the number itself. So, this is not a square number. Also, we know that,
\[16=4\times 4={{\left( 4 \right)}^{2}}\]
So, 16 is a product of a number multiplied by itself. So, it is a perfect square. Hence, 16 is a square number. We had assumed that m = 16. So, we can say that m is a square number but not a perfect number.
Hence, option (a) and (c) are correct answers.

Note: In this question, apart from m = 16, students can also take any other value of m which is natural and also equal to \[{{n}^{2}}\] where n is also a natural number like 4, 9, 25, 36, etc and follow the similar steps to arrive at the correct answer. Also, remember that there’s not any set formula for finding a perfect number, neither is there any pattern in perfect numbers but we just need to find their divisors and add them and see if it is equal to the original number or not.