A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m? Justify your answer.

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Solution

No, because by Euclid’s division of Lemma

b=aq + r

Here, b is any positive integer, a=3, b=3q+r where 0 ≤ r < 3

So this must be in the form of 3q, 3q+1, or 3q +2

Now, ${3 q}^{2}$ = ${9 q}^{2}$ = 3m [ where m= ${3 q}^{2}$ ]

And $(3 q + 1)^{2}$ = $(3 q)^{2}$ + 6q +1

= 3 $({3 q}^{2}$ +2q) + 1

= 3 m + 1 [m= $({3 q}^{2}$ + 2 q]

Also $(3 q + 2)^{2}$ = ${9 q}^{2}$ + 12 q + 4

= ${9 q}^{2}$ + 12 q +3 +1

= 3 $({3 q}^{2}$ + 4q + 1) +1

= 3 m +1 [ m = $({3 q}^{2}$ + 4q + 1) ]

Clearly from above, the expression could not be expressed in the form of either 3q or 3q + 2

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Similar questions

Q. A positive integer is of the form

3 q + 1, q

being a anatural number. Can you write its square in any form other than

3 m + 1, 3 m

3 m + 2

for some integer

m

? Justify your answer.

Question 5
A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1. i.e., 3m or 3m +2 for some integer m? Justify your answer.

Write whether the square of any positive integer can be of the form 3m+2 , where m is a natural number . Justify your answer ?

Question 4
Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.

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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