Show that the square of any positive integer cannot be of the form $3 m + 2$ , where $m$ is a natural number..

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Solution

By Euclid's lemma, $b = a q + r, 0 \leq r < a .$
Here, b is a positive integer and a = 3.
∴ $b = 3 q + r, for 0 \leq r < 3$
This must be in the form $3 q, 3 q + 1 or 3 q + 2$ .
Now,
${(3 q)}^{2} = 9 q^{2} = 3 m, where m = 3 q^{2} {(3 q + 1)}^{2} = 9 q^{2} + 6 q + 1 = 3 (3 q^{2} + 2 q) + 1 = 3 m + 1, where m = 3 q^{2} + 2 q {(3 q + 2)}^{2} = 9 q^{2} + 12 q + 4 = 3 (3 q^{2} + 4 q + 1) + 1 = 3 m + 1, where m = 3 q^{2} + 4 q + 1$
Therefore, the square of a positive integer cannot be of the form 3m + 2, where m is a natural number.

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Say true or false:

A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.

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

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

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Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number..

Show that the square of any positive integer cannot be of the form $3 m + 2$ , where $m$ is a natural number..