Q. 44.1( 25 Votes )

# Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Answer :

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer, then corresponding to the positive integers a and 6, there exist, non-negative integers q and r such that

a = 6q + r, where 0 ≤ r < 6

Squaring both the sides using (a + b)^{2} = a^{2} + 2ab + b^{2}

⇒ a^{2} = (6q + r)^{2} = 36q^{2} + r^{2} + 12qr

⇒ a^{2} = 6(6q^{2} + 2qr) + r^{2}

where,0 ≤ r < 6

**Case I** When r = 0 we get

⇒ a^{2} = 6(6q^{2}) = 6m

where, m = 6q^{2} is an integer.

**Case II** when r = 1 we get

⇒ a^{2} = 6(6q^{2} + 2q) + 1 = 6m + 1

where, m = (6q^{2} + 2q) is an integer.

**Case III** When r = 2 we get

⇒ a^{2} = 6(6q^{2} + 4q) + 4 = 6m + 4

where, m = (6q^{2} + 4q) is an integer.

**Case IV** When r = 3 we get

⇒ a^{2} = 6(6q^{2} + 6q) + 9

⇒ a^{2} = 6(6q^{2} + 6a) + 6 + 3

⇒ a^{2} = 6(6q^{2} + 6q + 1) + 3 = 6m + 3

where, m = (6q + 6q + 1) is an integer.

**Case V** when r = 4 we get

⇒ a^{2} = 6(6q^{2} + 8q) + 16

⇒ a^{2} = 6(6q^{2} + 10q) + 24 + 1

⇒ a^{2} = 6(6q^{2} + 8q + 2) + 4 = 6m + 4

where, m = (6q^{2} + 8q + 2) is an integer

**Case VI** When r = 5 we get

⇒ a^{2} = 6(6q^{2} + 10q) + 25

⇒ a^{2} = 6(6q^{2} + 10q) + 24 + 1

⇒ a^{2} = 6(6q^{2} + 10q + 4) + 1 = 6m + 1

where, m = (6q^{2} + 10q + 1) + 1 is an integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

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