Question

square of any positive integer can be in the form 3m+2

Answer 1

Let a be any positive integer. Then by Euclid’s division lemma, we have 
a = bq + r, where 0 ≤ r < b 
For b = 3, we have 
a = 3q + r, where 0 ≤ r < 3 ...(i)

So, The numbers are of the form 3q, 3q + 1 and 3q + 2. 
So, (3q)2 = 9q2 = 3(3q2) 
= 3m, where m is a integer. 
(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 
= 3m + 1, 
where m is a integer. 
(3q + 2)2 = 9q2 + 12q + 4, 
which cannot be expressed in the form 3m + 2. 
Therefore, Square of any positive integer cannot be expressed in the form 3m + 2.


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Answer 2

NO,by Euclid's lemma, b = aq + r,0 ≤ r ≤ a Here, b is any positive integer, a = 3b = 3q + r
for 0 ≤ r ≤ 2
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.