Question

Use euclid's lemma to show that the square of any positive integer is the author of the form of 3m or 3m + 1 for some integer m ?

Cläßß :- 10th

Chäp :- Rèàl Ñúmbèr

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Answer 1

Let a be any positive integer and b = 3.

Then, by Euclid's Division Lemma,

a = 3q + r where $0 \leqslant r < 3$.

r = 0, 1, or 2.

So, we have three cases.

$\textbf{Case-I}$

$a = 3q \\ \\ {a}^{2} = 9 {q}^{2} \\ \\ {a}^{2} = 3(3 {q}^{2} ) \\ \\ {a}^{2} = 3m\: \tt (where \: \it m \: \tt = 3 {q}^{2} )$

$\textbf {Case-II}$

$a = 3q + 1 \\ \\ {a}^{2} = {(3q + 1)}^{2} \\ \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ \\ {a}^{2} = 3(3 {q}^{2} + 2q) + 1 \\ \\ {a}^{2} = 3m + 1 \: \tt (where \: \it m \: \tt = 3 {q}^{2} + 2q)$

$\textbf {Case-III}$

$a = 3q + 2 \\ \\ {a}^{2} = {(3q + 2)}^{2} \\ \\ {a}^{2} = 9 {q}^{2} + 12q + 4 \\ \\ {a}^{2} = 9 {q}^{2} + 12q + 3 + 1 \\ \\ {a}^{2} = 3(3 {q}^{2} + 4q + 1) + 1 \\ \\ {a}^{2} = 3m + 1 \: \tt(where \: \it m \: \tt = 3 {q}^{2} + 4q + 1)$

Thus, the conclusions of the three cases justify that the square of any positive integer is in the form of $3m$ , or $(3m+1)$ for some integer $m$ .