# Show that the square of an odd integer can be of the form $6q+1$ or $6q+3$, for some integer $q$.

Given: Statement "Square of an odd integer can be of the form $6q+1$ or $6q+3$, for some integer $q$".

To do: Here we have to prove the given statement.

Solution:

According to Euclid's division lemma,

If $a$ and $b$ are two positive integers;

• $a\ =\ bm\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.

If $b\ =\ 6$, then;

• $a\ =\ 6m\ +\ r$, where $0\ \underline{< }\ r\ <\ 6$.
• But here we have to consider only odd positive integers.
• So, $r\ =\ 1,\ 3,\ 5$

When, $r\ =\ 1$:

$a\ =\ 6m\ +\ 1$

Squaring on both sides, we get:

$a^2\ = (6m\ +\ 1)^2$

$a^2\ = 36m^2\ +\ 12m\ + 1$

$a^2\ = 6(6m^2\ +\ 2m)\ +\ 1$

$a^2\ = 6q\ +\ 1$, where $q\ =\ 6m^2\ +\ 2m$

When, $r\ =\ 3$:

$a\ =\ 6q\ +\ 3$

Squaring on both sides, we get:

$a^2\ = (6m\ +\ 3)^2$

$a^2\ = 36m^2\ +\ 36m\ + 9$

$a^2\ = 36m^2\ +\ 36m\ + 6\ +\ 3$

$a^2\ = 6(6m^2\ +\ 6m\ +\ 1)\ +\ 3$

$a^2\ = 6q\ +\ 3$, where $q\ =\ 6m^2\ +\ 6m\ +\ 1$

When, $r\ =\ 5$:

$a\ =\ 6q\ +\ 5$

Squaring on both sides, we get:

$a^2\ = (6m\ +\ 5)^2$

$a^2\ = 36m^2\ +\ 60m\ + 25$

$a^2\ = 36m^2\ +\ 60m\ + 24\ +\ 1$

$a^2\ = 6(6m^2\ +\ 10m\ +\ 4)\ +\ 1$

$a^2\ = 6q\ +\ 1$, where $q\ =\ 6m^2\ +\ 10m\ +\ 4$

Hence, the square of an odd integer can be of the form $6q+1$ or $6q+3$, for some integer $q$.

Updated on: 10-Oct-2022

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