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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 $q$.