Question

Show that the square of any positive odd integer is of the form $4m+1$,for same integer $m$

Answer 1

To prove the square of any positive odd integer is of the form $4m+1$,for same integer $m$.

Euclid's Division Lemma: For any two positive integers $a$ and $b$, there exists unique integer $q$ and satisfying $a=bq+r$, where $0\le r<b$.

If $b=4$ then $a=4q+r$ where $0\le r<2$.

So, $r=0,1$.

Case 1: When $r=0$,

$\begin{array}{rcl}a& =& 4q+0\\ & =& 4q\end{array}$

Square on both sides.

$$\begin{gathered} a^2={\left(4q\right)}^2 \\ {a}^2=16q^2 \\ {a}^2=4m\mathrm{where},\left[m=4q^2\right] \end{gathered}$$

Case 2: When $r=1$,

$\begin{array}{rcl}a& =& 4q+1\end{array}$

Since, $4q+1$ is not divisible by $2$, thus, $4q+1$ is an odd integer.

Square on both sides.

$$\begin{gathered} a^2={\left(4q+1\right)}^2 \\ {a}^2=16q^2+1+8q \\ {a}^2=4\left(4q^2+2q\right)+1 \\ {a}^2=4m+1\mathrm{where},\left[m=4q^2+2q\right] \end{gathered}$$

Hence, it is proved that the square of any odd integer is of the form $4m+1$, for same integer $m$.