# Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m$+$1 for some integer m.

Given :

The given positive integer is 'm'.

To do :

We have to show that the square of any positive integer is either of the form 3m or 3m$+$1 for some integer m.

Solution :

By Euclid's division algorithm,

$a = b q + r$, where $0 \leq r < b$.

Let a be the positive integer and b $=$ 3.

Now, $a = 3 q + r$, where $0 \leq r < 3$.

The possibilities of r is 0 , 1 , 2.

When $r = 0$,

$a = 3 q + r$

$a = 3 q + 0$

$a = 3 q$

Squaring on both sides,

$a^2 = (3 q)^2$

$a^2 = 9 q^2$

$a^2 = 3(3 q^2)$

$a^2 = 3 m$ ; where $m = 3 q^2$.

When $r = 1$

$a = 3 q + r$

$a = 3 q + 1$

Squaring on both sides,

$a^2 = (3 q + 1)^2$

$a^2 = (3 q)^2 + 1^2 + 2 (3q) (1)$                           $[(a+b)^2 = a^2 + b^2 + 2ab]$

$a^2 = 9 q^2 + 1 + 6q$

$a^2 = 9 q^2 + 6q + 1$

$a^2 = 3(3 q^2 + 2 q) + 1$

$a^2 = 3 m + 1$ ; where $m = 3 q^2 + 2 q$.

When $r = 2$

$a = 3 q + r$

$a = 3 q + 2$

Squaring on both sides,

$a^2 = (3 q + 2)^2$

$a^2 = (3 q)^2 + 2^2 + 2 (3q) (2)$                           $[(a+b)^2 = a^2 + b^2 + 2ab]$

$a^2 = 9 q^2 + 4 + 12q$

$a^2 = 9 q^2 + 12 q + 4$

$a^2 = 9 q^2 + 12 q + 3 + 1$

$a^2 = 3(3 q^2 + 4 q + 1) + 1$

$a^2 = 3 m + 1$ ; where $m = 3 q^2 + 4 q + 1$.

Hence, the square of any positive integer is either of the form 3m or 3m$+$1 for

some integer m.

Updated on: 10-Oct-2022

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