Prove that square of any positive integer is always of the form $4 m, 4 m + 1, 4 m + 4$ or $4 m + 9$ .

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Solution

Let $a$ be any positive integer and $b = 4$
then by Euclid division lemma
$a = 4 q + r$ where $r = 0, 1, 2, 3$
$a = 4 q, a = 4 q + 1, a = 4 q + 2, a = 4 q + 3$
If,
$a = 4 q, a^{2} = (4 q)^{2} = 16 q^{2} = 4 (4 q^{2}) = 4 m$ where $m = 4 q^{2}$
$a = 4 q + 1, a^{2} = (4 q + 1)^{2} = 16 q^{2} + 1 + 8 q = 4 (4 q^{2} + 2 q) + 1 = 4 m + 1$ where $m = 4 q^{2} + 2 q$
$a = 4 q + 2, a^{2} = (4 q + 2)^{2} = 16 q^{2} + 4 + 16 q = 4 (4 q^{2} + 4 q) + 4 = 4 m + 4$ where $m = 4 q^{2} + 4 q$
$a = 4 q + 3, a^{2} = (4 q + 3)^{2} = 16 q^{2} + 9 + 24 q = 4 (4 q^{2} + 6 q) + 9 = 4 m + 9$ where $m = 4 q^{2} + 6 q$
Hence, square of any positive integer is of the form $4 m, 4 m + 1, 4 m + 4, 4 m + 9$
Where $m$ is any integer.

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