**Solution:**

Let a be the positive integer and b = 5.

Then, by Euclid’s algorithm, a = 5m + r for some integer m ≥ 0 and r = 0, 1, 2, 3, 4 because 0 ≤ r < 5.

So, a = 5m or 5m + 1 or 5m + 2 or 5m + 3 or 5m + 4.

So, (5m)^{2} = 25m^{2} = 5(5m^{2})

= 5q, where q is any integer.

(5m + 1)^{2} = 25m^{2} + 10m + 1

= 5(5m^{2} + 2m) + 1

= 5q + 1, where q is any integer.

(5m + 2)^{2} = 25m^{2} + 20m + 4

= 5(5m^{2} + 4m) + 4

= 5q + 4, where q is any integer.

(5m + 3)^{2} = 25m^{2} + 30m + 9

= 5(5m^{2} + 6m + 1) + 4

= 5q + 4, where q is any integer.

(5m + 4)^{2} = 25m^{2} + 40m + 16

= 5(5m^{2} + 8m + 3) + 1

= 5q + 1, where q is any integer.

**Hence, The square of any positive integer is of the form 5q, 5q + 1, 5q + 4 and cannot be of the form 5q + 2 or 5q + 3 for any integer q.**