Solution

By Euclid's division algorithm , $$ a = bq + r $$ where $$ a , b , q , r $$ are non-negative integers and $$ 0 \leq r < b$$.
On putting $$ b = 4 $$ we get
$$ a = 4q + r $$ ....(i)

When $$ r = 0 , a = 4q $$ which is even (as it is divisible by $$2$$)

When $$ r = 1 , a = 4q + 1 $$ which is odd (as it is not divisible by $$2$$)
Squaring the odd number $$ (4q + 1) $$ , we get
$$a^2= (4q + 1)^2 $$
$$ = (4q)^2 + (1)^2 + 2(4q) $$
$$ = 4[4q^2 + 2q ] + 1 $$
$$ = 4m + 1 $$ is perfect square for $$ m = 4q^2 + 2q $$

When $$ r = 2 , a = 4q + 2 $$ [From(i)]
$$ \Rightarrow \,a = 2(2q + 1) $$ which is even (as it is divisible by $$2$$)


When $$ r = 3 , a = 4q + 3 = 4q + 2 + 1 $$
$$ = 2[2q + 1] + 1 $$ which is odd (as it is not divisible by $$2$$)
Squaring the odd number $$ (4q + 3) $$ , we get
$$ a^2=(4q + 3)^2 = (4q)^2 + (3)^2 + 2(4q) (3) $$
$$ = 16q^2 + 9 + 24q $$
$$ = 16q^2 + 24q + 8 + 1 $$
$$ = 4[4q^2 \,6q + 2] + 1 $$
$$ = 4m + 1 $$ is perfect square for some value of $$m$$.

Hence, the square on any odd integer is of the form $$(4q+1)$$ for some integer $$q$$.