Show that the square of any positive integer is either of the form $4 q$ or $4 q + 1$ for some integer $q$ .

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To prove the square of any positive integer is either of the form $4 q$ or $4 q + 1$ for some integer $q$ .
Euclid's Division Lemma: For any two positive integers $a$ and $b$ , there exists unique integer $m$ and satisfying $a = b m + r$ , where $0 \leq r < b$ .
If $b = 4$ then $a = 4 m + r$ where $0 \leq r < 4$ .
So, $r = 0, 1, 2, 3$ .
Case 1: When $r = 0$ ,
$\begin{array}{rcl} a & = & 4 m + 0 \\ = & 4 m \end{array}$
Square both sides.
$a^{2} = {(4 m)}^{2} a^{2} = 16 m^{2} a^{2} = 4 q where, [q = 4 m^{2}]$
Case 2: When $r = 1$ ,
$\begin{array}{rcl} a & = & 4 m + 1 \end{array}$
Square both sides.
$a^{2} = {(4 m + 1)}^{2} a^{2} = 16 m^{2} + 1 + 8 m a^{2} = 4 (4 m^{2} + 2 m) + 1 a^{2} = 4 q + 1 where, [q = 4 m^{2} + 2 m]$
Case 3: When $r = 2$ ,
$\begin{array}{rcl} a & = & 4 m + 2 \end{array}$
Square both sides.
$a^{2} = {(4 m + 2)}^{2} a^{2} = 16 m^{2} + 4 + 16 m a^{2} = 4 (4 m^{2} + 4 m + 1) a^{2} = 4 q where, [q = 4 m^{2} + 4 m + 1]$
Case 4: When $r = 3$ ,
$\begin{array}{rcl} a & = & 4 m + 3 \end{array}$
Square both sides.
$a^{2} = {(4 m + 3)}^{2} a^{2} = 16 m^{2} + 9 + 24 m a^{2} = 4 (4 m^{2} + 6 m + 2) + 1 a^{2} = 4 q + 1 where, [q = 4 m^{2} + 6 m + 2]$
Hence, it is shown that the square of any positive integer is either of the form $4 q$ or $4 q + 1$ for some integer $q$ .

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