Question

Prove that every even power of every odd number is of the form $ 8r+1 $ .

Answer 1

Hint: We first try to form the general equation form for even and odd numbers. We then use the form to find the power of the odd number. We try to express it in the form of $ 8r+1 $ .

Complete step by step solution:
We have to prove that every even power of every odd number is of the form $ 8r+1 $ .
Every odd number is of the form $ 2k+1,k\in \mathbb{Z} $ whereas the even numbers are of the form $ 2k,k\in \mathbb{Z} $ .
We take even power of an odd number.
Let the odd number be $ p=2m+1 $ for some $ m\in \mathbb{Z} $ .
We take $ {{p}^{2k}}={{\left( {{p}^{2}} \right)}^{k}} $ . Now we try to find the value of the $ {{p}^{2}} $ .
So,
 $ \Rightarrow {{p}^{2}}={{\left( 2m+1 \right)}^{2}}=4{{m}^{2}}+4m+1 $ .
We applied $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ .
For the equation $ {{p}^{2}}=4{{m}^{2}}+4m+1 $ , we take common term $ 4m $ and get
 $ \Rightarrow {{p}^{2}}=4{{m}^{2}}+4m+1=4m\left( m+1 \right)+1 $ .
Therefore, we get multiplication of two numbers $ m $ and $ \left( m+1 \right) $ . As they are consecutive, one of them has to be even. Therefore, we can take 2 common from the multiplication of $ m\left( m+1 \right) $ .
Therefore,
 $ {{p}^{2}}=4m\left( m+1 \right)+1=8r+1 $ where $ \dfrac{m\left( m+1 \right)}{2}=r $ .
Thus proved, every even power of every odd number is of the form $ 8r+1 $ .

Note: We can also show that any power form of the form $ 8r+1 $ , be it odd or even always remains as the form of $ 8r+1 $ . We can always use the concept of induction to prove it.