Question

a positive integer is of the form 3q + 1 q being a natural number can you write its square in any form other than 3m+1 that it 3m 3m+ 2 for some integer m justify your answer​

Answer 1

Answer:

No.

Step-by-step explanation:

No, by Eucblid's Lemma,b = aq + r,0 ≤ r < a

Here, b is any positive integer a = 3, b = 3 q + r for 0 ≤ r < 3

So, this must be in the form 3q, 3q+1 or 3 q+2.

Now, (3q)² = 9q²= 3m [here, m = 3q²]

and (3q+1)² = 9 q²+ 6q + 1

= (3q+1)² = 9q² + 6q + 1

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

Also, (3q+2)² = 9q² + 12q + 4

= 9 q² + 12 q + 3 + 1

= 3(3 q² + 4 q + 1) + 1

= 3m + 1 [here,m = 3 q² + 4q + 1]

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.