Question

show that the square any positive integer is of the form 3M or 3M + 1 for some integer m

Answer 1

Hope this answer will help you.

Answer 2

$\huge \fbox \red{Solution:}$

No,

Justification:

By Euclid's Division Lemma,

a = bq + r, 0 ≤ r < b

Here, a is any positive integer and b = 3,

⇒ a = 3q + r

So, a can be of the form 3q, 3q + 1 or 3q + 2.

Now, for a = 3q(3q)2

= 3(3q2) = 3m [where m = 3q2]

for a = 3q + 1

(3q + 1)2

= 9q2+ 6q + 1 = 3(3q2+ 2q) + 1 = 3m + 1 [where m = 3q2+2q]

for a = 3q + 2(3q + 2)2

= 9q2+ 12q + 4 = 9q2+ 12q + 3 + 1 = 3(3q2+ 4q + 1) + 1

= 3m + 1 [where m = 3q2+ 4q + 1]

Thus, square of a positive integer of the form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.