Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.

Solution:

To Prove: that the square of an odd positive integer is of the form 8q + 1, for some integer q.

Proof: Since any positive integer n is of the form 4m + 1 and 4m + 3

If n = 4m + 1

$\Rightarrow n^{2}=(4 m+1)^{2}$

$\Rightarrow n^{2}=(4 m)^{2}+8 m+1$

$\Rightarrow n^{2}=16 m^{2}+8 m+1$

$\Rightarrow n^{2}=8 m(2 m+1)+1$

$\Rightarrow n^{2}+8 q+1(q=m(2 m+1))$

If n = 4m + 3

$\Rightarrow n^{2}=(4 m+3)^{2}$

$\Rightarrow n^{2}=(4 m)^{2}+24 m+9$

$\Rightarrow n^{2}=16 m^{2}+24 m+9$

$\Rightarrow n^{2}=8\left(2 m^{2}+3 m+1\right)+1$

$\Rightarrow n^{2}=8 q+1 \quad\left(q=\left(2 m^{2}+3 m+1\right)\right)$

Hence n2 integer is of the form 8q + 1, for some integer q.