Question

Question 5
Show that the square of any odd integer is of the form 4m + 1, for some integer m.

Answer 1

By Euclid's division algorithm, we have $a=bq+r,where0\le r<b$
On putting b = 4, we get,
$a=4q+rwhere0\le r<4$ i.e., r = 0, 1, 2, 3
If r = 0 $\Rightarrow$ a = 4q is divisible by 2 $\Rightarrow$ 4q is even.
If $r=2\Rightarrow a=4q+2,2(2q+1)$ is divisible b y 2 $\Rightarrow$ 2 (2q + 1) is even.
If $r=3\Rightarrow a=4q+3,(4q+1)and(4q+3)$ are odd integers.
So, for any positive integer q, (4q + 1) and (4q + 3) are odd integers.
Now, $a^2=(4q+1{)}^2=16q^2+1+8q=4(4q^2+2q)+1$
$[\because (a+b{)}^2=a^2+2ab+b^2]$
This is a square which is of the form 4m + 1, where $m=(4q^2+2q)$ is an integer.
and $a^2=(4q+3{)}^2=16q^2+9+24q=4(4q^2+6q+2)+1$ is a square.
$[\because (a+b{)}^2=a^2+2ab+b^2]$
Which is of the form 4m + 1, where m = $(4q^2+6q+2)$ is an integer.
Hence, for some integer m, the square of any odd integer is of the form 4m + 1.