Show that the square of any positive integer never leaves a remainder 5 or 6 when divided by 7.

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Solution

Let $a$ be any positive integer then for $a a n d 7$ by Euclid's division lemma we have,
$a = 7 q + r$ , where $r = 0, 1, 2, 3, 4, 5, 6$

Case 1: when $r = 0$
$a = 7 q$

$\begin{matrix} a^{2} = 49 q^{2} C a s e 2 : W h e n r = 1 a = 7 q + 1 a^{2} = {(7 q + 1)}^{2} = 49 q^{2} + 1 + 14 q = 7 (7 q^{2} + 2 q) + 1 C a s e 3 : W h e n r = 2 a = 7 q + 2 a^{2} = {(7 q + 2)}^{2} = 49 q^{2} + 4 + 28 q a^{2} = 7 (7 q^{2} + 4 q) + 4 C a s e 4 : W h e n r = 3 a = 7 q + 3 a^{2} = {(7 q + 3)}^{2} = 49 q^{2} + 9 + 42 q a^{2} = 7 (7 q^{2} + 6 q + 1) + 2 C a s e 5 : W h e n r = 4 a = 7 q + 4 a^{2} = {(7 q + 4)}^{2} = 49 q^{2} + 16 + 56 q a^{2} = 7 (7 q^{2} + 8 q + 2) + 2 C a s e 6 : W h e n r = 5 a = 7 q + 5 a^{2} = {(7 q + 5)}^{2} = 49 q^{2} + 25 + 70 q a^{2} = 7 (7 q^{2} + 10 q + 3) + 4 C a s e 7 : W h e n r = 6 a = 7 q + 6 a^{2} = {(7 q + 6)}^{2} = 49 q^{2} + 36 + 84 q a^{2} = 7 (7 q^{2} + 12 q + 5) + 1 \end{matrix}$
Hence, from case 1 to case 7 we can say that when the square of any positive integer is divided by $7$ then it never leaves $5 o r 6$

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