Question

Please help.............. Show that the square of any positive integer can be of the form 3m+2 where m is a natural Number. Justify your answer

Answer 1

Hello there !!
Here is your Answer !!

NO,
by Euclid's lemma,
b = aq + r,
Here, b is any positive integer,
a = 3b = 3q + r

So, any positive integer is of the form 3k, 3k + 1 or 3k + 2.

Now, (3k)2 = 9k2 = 3m [where, m = 3k2]
and (3k + 1)2 = 9k2 + 6k + 1
= 3(3k2 + 2k) + 1 = 3m + 1[where, m = 3k2+ 2k]
Also, (3k+2)2 = 9k2 + 12k + 4[∵(a+b)2 = a2 + 2ab + b2]
= 9k2 + 12k + 3 + 1
= 3(3k2 + 4k + 1) + 1
= 3m + 1 [where, m = 3k2 + 2k]

which is in the form of 3m and 3m + 1.
Hence, square of any positive number cannot be of the form 3m + 2.     

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