Question

write whether the square of any positive integer can be of the form 3m+2, where m is is a a natural number. justify your answer.​

Answer 1

Answer:

${\huge{\boxed{\mathcal{\blue{✍︎✍︎ Bᴇsᴛ Aɴsᴡᴇʀ:-}}}}}$

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.

Answer 2

Answer:

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.