Question

Show that the square of any odd integer is of the form 4q+1, for some integer q.

Answer 1

Hint: We solve this question by using the method of Mathematical induction. First, we check if the given statement is true for the number 1. Then we assume that the statement is true for some odd integer k, then we find if the statement is true for the immediate next odd integer k+2. If it is true then we say that the given statement is true.

Complete step by step answer:
Let us prove this using the Proof by Mathematical Induction.
Here the given statement is square of any odd is of the form 4q+1, for some integer q.
First, let us consider the first odd integer 1.
Square of 1 is 1.
We can write 1 as,
$1=4\left( 0 \right)+1$
So, here q=0. As 0 is an integer, the given statement is true for odd integer 1.
Now let us assume that the given statement is true for an integer k which is odd, that is
$\Rightarrow {{k}^{2}}=4q+1$
The immediate odd integer to k is k+2.
Now let us check the statement for the immediate odd integer k+2.
Then the square of the number k+2 is,
\[\Rightarrow {{\left( k+2 \right)}^{2}}={{k}^{2}}+4k+4\]
Now, let us substitute the value of ${{k}^{2}}$ obtained above in it.
\[\begin{align}
  & \Rightarrow {{\left( k+2 \right)}^{2}}=4q+1+4k+4 \\
 & \Rightarrow {{\left( k+2 \right)}^{2}}=4q+4k+4+1 \\
 & \Rightarrow {{\left( k+2 \right)}^{2}}=4\left( q+k+1 \right)+1 \\
 & \Rightarrow {{\left( k+2 \right)}^{2}}=4Q+1 \\
\end{align}\]
So, the statement is true for k+2.
That is if the statement is true for the odd integer k, then it is also true for the immediate next odd integer.
So, by Mathematical Induction we can say that the given statement is true.
Hence, the square of any odd integer is of the form 4q+1, for some integer q.
Hence Proved.

Note: We can also solve this question in an alternative process.
Here we need to prove that square of any odd integer is of the form 4q+1, for some integer q.
As we know, any odd integer can be written in the form of 2n+1.
Now, let us consider the square of this integer.
$\begin{align}
  & \Rightarrow {{\left( 2n+1 \right)}^{2}}=4{{n}^{2}}+4n+1 \\
 & \Rightarrow {{\left( 2n+1 \right)}^{2}}=4\left( {{n}^{2}}+n \right)+1 \\
 & \Rightarrow {{\left( 2n+1 \right)}^{2}}=4q+1 \\
\end{align}$
So, we can write squares of any odd integer of the form 4q+1, for some integer q.
Hence Proved.