**Solution:**

Let a be any odd integer and b = 4.

Then, by Euclid’s algorithm, a = 4m + r for some integer m ≥ 0 and r = 0,1,2,3 because 0 ≤ r < 4.

So, a = 4m or 4m + 1 or 4m + 2 or 4m + 3

So, a = 4m + 1 or 4m + 3

Here, a cannot be 4m or 4m + 2, as they are divisible by 2.

=> (4m + 1)^{2} = 16m^{2} + 8m + 1

= 4(4m^{2} + 2m) + 1

= 4q + 1, where q is some integer.

(4m + 3)^{2} = 16m^{2} + 24m + 9

= 4(4m^{2} + 6m + 2) + 1

= 4q + 1, where q is some integer.

**Hence, The square of any odd integer is of the form 4q + 1, for some integer q.**