Question

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

Answer 1

Answer:

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Answer 2

Answer:

As per the Euclid's division Lemma

If a and b are two positive integers

$a = bq + r$

$where \: 0 < a < b$

$let \: positive \: integer \: a \: and \: b = 3$

$hence \: a \: = 3q + r$

$r \: is \: an \: integer \: greater \: than \: or \: equal \: to \: 0 \: but \: less \: than \: 3$

$case \: 1 \\ if \: r = 0$

$a = 3q + r \\ a = 3q + 0 \\ a = 3q \\ squaring \: both \: sides \: we \: get \\ a {}^{2} = 3q {}^{2} \\ a {}^{2} = 9q {}^{2} \\ a {}^{2} = 3(3q) {}^{2} \\ where \: m = 3q {}^{2}$

$case \: 2 \\ if \: r = 1 \\ a = 3q + r \\ a = 3q + 1 \\ squaring \: both \: sides \: we \: get \\ a {}^{2} = (3q + 1) {}^{2} \\ a {}^{2} = 3q {}^{2} + 2 \times 3q \times 1 + 1 {}^{2} \\ a {}^{2} = 9q {}^{2} + 6q + 1 \\ a {}^{2} = 3(3q {}^{2} + 2q) + 1 \\ where \: m = 3q {}^{2} + 2q$

$case \: 3 \\ if \: r = 2 \\ a = 3q + r \\ a = 3q + 2 \\ squaring \: both \: sides \: we \: get \\ a {}^{2} = (3q + 2) {}^{2} \\ a {}^{2} = 3q {}^{2} + 2 \times 3q \times 2 \times 2 {}^{2} \\ a {}^{2} = 9q {}^{2} + 12q + 4 \\ a {}^{2} = 3(3q {}^{2} + 4q + 1) + 1 \\ where \: m = 3q {}^{2} + 4q + 1$

Hence square of any positive integer can be expressed in the form 3m,3m+1 or 3m+2.

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