Question

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

Answer 1

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

Apply Euclid's division algorithm, $a=bm+r$ where $a,b,m,r$ are non-negative integers and $0\le r<b$.

Assume that $b=4$.

Substitute $r=0,1,2,3$ and $b=4$ into $a=bm+r$.

$\begin{array}{rcl}a& =& 4m\left[\therefore r=0\right]\\ a& =& 4m+1\left[\therefore r=1\right]\\ a& =& 4m+2\left[\therefore r=2\right]\\ a& =& 4m+3\left[\therefore r=3\right]\end{array}$

We know that $a$ is odd

So, the value of $a$ cannot be $4m\mathrm{and}4m+2$ since $4m\mathrm{and}4m+2$ are divisible by $2$ and so are even.

Thus, $a=4m+1\mathrm{and}a=4m+3$.

Square $a=4m+1$ on both sides.

$\begin{array}{rcl}{a}^2& =& {\left(4m+1\right)}^2\\ & \Rightarrow & a^2=16m^2+8m+1\\ & \Rightarrow & a^2=4\left(4m^2+2m\right)+1\\ & \Rightarrow & a^2=4q+1\left[q\mathrm{is}\mathrm{some}\mathrm{integer}\mathrm{and}q=4m^2+2m\right]\end{array}$

Square $a=4m+3$ on both sides.

$\begin{array}{rcl}{a}^2& =& {\left(4m+3\right)}^2\\ & \Rightarrow & a^2=16m^2+24m+9\\ & \Rightarrow & a^2=4\left(4m^2+6m+2\right)+1\\ & \Rightarrow & a^2=4q+1\left[q\mathrm{is}\mathrm{some}\mathrm{integer}\mathrm{and}q=4m^2+6m+2\right]\end{array}$

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