Question

Show that the difference between any number and its square is even.

Answer 1

Hint: This problem deals with multiplication of even number and odd number, odd number and even number. The product of an even number and another even number always results in an even number, but whereas the product of an even number and an odd number always results in an odd number.

Complete step-by-step answer:
Given any number, and the square of the number.
We have to prove that the difference between the number and the square of the number is an even number.
So let the number be \['p'\]
Now consider the number $ p $ , and square the number.
The square of the number $ p $ is given by:
 $ \Rightarrow {\left( p \right)^2} = {p^2} $
Now the difference of the number $ p $ and the square of the number $ p $ is given below:
 $ \Rightarrow p - {p^2} $
Here taking the number $ p $ common from the above expression, as given below:
 $ \Rightarrow p\left( {1 - p} \right) $
Hence the difference between any number $ p $ and its square is $ p\left( {1 - p} \right) $ .
Here there are cases, when $ p $ is even and when $ p $ is odd.
When $ p $ is even,
If $ p $ is an even number, then $ p - 1 $ would be an odd number.
The product of an even number and odd number always results in an even number.
Hence the product $ p\left( {1 - p} \right) $ is an even number.
When p is odd,
If $ p $ is an odd number, then $ p - 1 $ would be an even number.
The product of an odd number and even number always results in an even number.
Hence the product $ p\left( {1 - p} \right) $ is an even number.
Thus in both the cases the product is an even number.
 $ \therefore $ The difference between any number and its square is even.

Hence proved.

Note:
While solving this problem we need to understand that whatever number we consider and subtract one from the number, these two numbers are always consecutive numbers, that is these numbers reside each other. Hence the product of the consecutive numbers is always even.