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

(N/A) Let $a$ be an odd integer. By Euclid's division lemma,for any positive integer $b=4$,we have $a = 4k + r$,where $0 \leq r < 4$.
Since $a$ is odd,$r$ can only be $1$ or $3$.
Case $1$: If $r = 1$,then $a = 4k + 1$.
$a^2 = (4k + 1)^2 = 16k^2 + 8k + 1 = 4(4k^2 + 2k) + 1$.
Let $q = 4k^2 + 2k$,which is an integer. Thus,$a^2 = 4q + 1$.
Case $2$: If $r = 3$,then $a = 4k + 3$.
$a^2 = (4k + 3)^2 = 16k^2 + 24k + 9 = 16k^2 + 24k + 8 + 1 = 4(4k^2 + 6k + 2) + 1$.
Let $q = 4k^2 + 6k + 2$,which is an integer. Thus,$a^2 = 4q + 1$.
In both cases,the square of an odd integer is of the form $4q + 1$ for some integer $q$.