Question

$\huge\mathfrak\purple{Heya\:Mate}$❤

Show that the square of a positive integer cannot be of the form 5q + 2 and 5q + 3 for an integer q. ​

Answer 1

Consider first a few perfect squares.

First 10 perfect squares are given below.

1, 4, 9, 16, 25, 36, 49, 64, 81, 100,,...

Divide each perfect square by 5 and write down each remainder thus formed.

1, 4, 4, 1, 0, 1, 4, 4, 1, 0,...

Here we can see that the sequence formed by the remainders is the repetition of the sequence 1, 4, 4, 1, 0.

There are only remainders 0, 1 and 4 appeared in this sequence, but not 2 and 3.

From this we get that the perfect squares leave either of 0, 1, and 4 as remainder on division by 5, but not 2 and 3.

Perfect squares which leave remainder 0 on division by 5 are in the form 5q.

Perfect squares which leave remainder 1 on division by 5 are in the form 5q + 1.

Perfect squares which leave remainder 4 on division by 5 are in the form 5q + 4.

As there is no perfect square which leaves remainder 2 or 3 on division by 5, they are not in the form either 5q + 2 or 5q + 3.

Hence proved!!!

Answer 2

To show : The the square of a positive integer cannot be of the form 5q + 2 and 5q + 3 for an integer q.

Solution:

Let 'a' be any positive integer and b = 5.

Using Euclid's Division Lemma -

a = bm + r [ 0 ≤ r < b ]

⇒ a = 5m + r [0 ≤ r < 5 ]

Now, possible values of r = 0,1,2,3,4.

For r = 0,

⇒ a = 5m

Squaring both sides -

⇒ a² = (5m)²

⇒ a² = 25m²

⇒ a² = 5 (5m²)

⇒ a² = 5q, where q = 5m²

For r = 1,

⇒ a = 5m + 1

Squaring both sides -

⇒ a² = (5m + 1)²

⇒ a² = 25m² + 1² + 2 * 5m * 1

⇒ a² = 25m² + 10m + 1

⇒ a² = 5 (5m² + 2m) + 1

⇒ a² = 5q + 1, where q = 5m² + 2m

For r = 2,

⇒ a = 5m + 2

Squaring both sides -

⇒ a² = (5m + 2)²

⇒ a² = 25m² + 2² + 2 * 5m * 2

⇒ a² = 25m² + 20m + 4

⇒ a² = 5 (5m² + 4m) + 4

⇒ a² = 5q + 4, where q = 5m² + 4m

For r = 3,

⇒ a = 5m + 3

Squaring both sides -

⇒ a² = (5m + 3)²

⇒ a² = 25m² + 3² + 2 * 5m * 3

⇒ a² = 25m² + 30m + 9

⇒ a² = 5 (5m² + 6m) + 9

⇒ a² = 5q + 9, where q = 5m² + 6m

For r = 4,

⇒ a = 5m + 4

Squaring both sides -

⇒ a² = (5m + 4)²

⇒ a² = 25m² + 4² + 2 * 5m * 4

⇒ a² = 25m² + 40m + 15 + 1

⇒ a² = 5 (5m² + 8m + 3) + 1

⇒ a² = 5q + 1, where q = 5m² + 8m + 3

From all above situations, it is clear that the square of any positive integer cannot be of the form 5q + 2 and 5q + 3 for an integer q.