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Prove that the square of any positive integer is of the form $4q$ or $4q+1$ for some integer q.
Given: Positive integer $q$.
To prove: We have to prove that square of any positive integer is of the form $4q$ or $4q\ +\ 1$ for some integer '$q$'.
Solution:
By Euclid's division algorithm,
If $a$ and $b$ are two positive integers,
$a\ =\ bm\ +\ r$, where $0\ \underline{< }\ r\ <\ b$
Let $a$ be the positive integer and $b$ is equal to 4,
$a\ =\ 4m\ +\ r$, where $0\ \underline{< }\ r\ <\ 4$,
So, $r\ =\ 0,\ 1,\ 2,\ 3$
Now,
When, $r\ =\ 0$,
$a\ =\ 4m$
Squaring on both sides, we get:
$a^{2}\ =\ (4m)^{2}$
$a^{2}\ =\ 4(4m^{2})$
$a^{2}\ =\ 4q$ , where $q\ =\ 4m^{2}$
When, $r\ =\ 1$,
$a\ =\ 4m\ +\ 1$
Squaring on both sides, we get:
$a^{2}\ =\ (4m\ +\ 1)^{2}$
$a^{2}\ =\ 16m^{2}\ +\ 1\ +\ 8m$
$a^{2}\ =\ 4(4m^{2}\ +\ 2m)\ +\ 1$
$a^{2}\ =\ 4q\ +\ 1$, where $q\ =\ 4m^{2}\ +\ 2m$
When, $r\ =\ 2$,
$a\ =\ 4m\ +\ 2$
Squaring on both sides, we get:
$a^{2}\ =\ (4m\ +\ 2)^{2}$
$a^{2}\ =\ 16m^{2}\ +\ 4\ +\ 16m$
$a^{2}\ =\ 4(4m^{2}\ +\ 4m\ +\ 1)$
$a^{2}\ =\ 4q$, where $q\ =\ 4m^{2}\ +\ 4m\ +\ 1$
When, $r\ =\ 3$,
$a\ =\ 4m\ +\ 3$
Squaring on both sides, we get:
$a^{2}\ =\ (4m\ +\ 3)^{2}$
$a^{2}\ =\ 16m^{2}\ +\ 9\ +\ 24m$
$a^{2}\ =\ 16m^{2}\ +\ 24m\ +\ 8\ +\ 1$
$a^{2}\ =\ 4(4m^{2}\ +\ 6m\ +\ 2)\ +\ 1$
$a^{2}\ =\ 4q\ +\ 1$ , where $q\ =\ 4m^{2}\ +\ 6m\ +\ 2$
Hence, the square of any positive integer is in the form of 4q or $4q\ +\ 1$, where $q$ is an integer.
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