Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

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 = 25m² = 5(5m²)
= 5q, where q is any integer.
(5m + 1)² = 25m² + 10m + 1
= 5(5m² + 2m) + 1
= 5q + 1, where q is any integer.
(5m + 2)² = 25m² + 20m + 4
= 5(5m² + 4m) + 4
= 5q + 4, where q is any integer.
(5m + 3)² = 25m² + 30m + 9
= 5(5m² + 6m + 1) + 4
= 5q + 4, where q is any integer.
(5m + 4)² = 25m² + 40m + 16
= 5(5m² + 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.

Let the positive integer = a

According to Euclid’s division lemma,

a = bm + r

According to the question, b = 5

a = 5m + r

So, r= 0, 1, 2, 3, 4

When r = 0, a = 5m.

When r = 1, a = 5m + 1.

When r = 2, a = 5m + 2.

When r = 3, a = 5m + 3.

When r = 4, a = 5m + 4.

Now,

When a = 5m

a² = (5m)² = 25m²

a² = 5(5m²) = 5q, where q = 5m²

When a = 5m + 1

a² = (5m + 1)² = 25m² + 10 m + 1

a² = 5 (5m² + 2m) + 1 = 5q + 1, where q = 5m² + 2m

When a = 5m + 2

a² = (5m + 2)²

a² = 25m² + 20m + 4

a² = 5 (5m² + 4m) + 4

a² = 5q + 4 where q = 5m² + 4m

When a = 5m + 3

a² = (5m + 3)² = 25m² + 30m + 9

a² = 5 (5m² + 6m + 1) + 4

a² = 5q + 4 where q = 5m² + 6m + 1

When a = 5m + 4

a² = (5m + 4)² = 25m² + 40m + 16

a² = 5 (5m² + 8m + 3) + 1

a² = 5q + 1 where q = 5m² + 8m + 3

Therefore, square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Hence Proved.