Question

show that the square of any positive integer cannot be of the form 3M + 2, where n is a natural number.​

Answer 1

Justification:
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.