Use Euclid's division lemma to show that the square of any positive integer is either of the form $3 m$ or $3 m + 1$ for some integer $m$ .

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Solution

From Euclid's division lemma
$x = 3 K + r; 0 \leq r < 2$
Case $(1); r = 0$
$\Rightarrow x = 3 K$
$ $\Rightarrow x^{2} = 9 K^{2} = 3 (3 K^{2}) = 3 M$

Case $(2) r = 1$
$\Rightarrow x = 3 K + 1$
$\Rightarrow x^{2} = 9 K^{2} + 6 K + 1$
$= 3 (3 K^{2} + 2 K) + 1$
$= 3 M + 1$

Case $(3) r = 2$
$\Rightarrow x = 3 K + 2$
$\Rightarrow x^{2} = 9 K^{2} + 12 K + 4$
$= 3 (3 K^{2} + 4 K + 1) + 1$
$= 3 M + 1$
$∴$ square of any positive integer is either of from $3 m$ or $3 M + 1$ for same possible inter $M$ . .

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