Question

Show that the square of any odd positive integer is of the form $8q+1$, where $q$ is a positive integer.

Answer 1

Let the odd positive integer be $2k+1$

${(2k+1)}^2=4k^2+4k+1$

$=4(k^2+k)+1$

$=4\left(k\right(k+1\left)\right)+1$

If k is odd, $(k+1)$ is even

If k is even, $(k+1)$ is odd

$\therefore k(k+1)$ is always even$=2q$ (say)

$\therefore {(2k+1)}^2=4.2q+1$

$=8q+1$