Question

use euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m (hint: let x be any positive integer then it is of from 3q,3q+1 or 3q+2 .Now square each of these and show that they can be rewritten in the form 3m or 3m + 1​

Answer 1

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1]

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Solution:

Let's consider a positive integer ‘a’.

By Euclid’s division lemma, we know that for any two positive integers a and b, there exist unique integers q and r, such that a = bq + r, 0 ≤ r < b

Let b = 3, then 0 ≤ r < 3. So, r = 0 or 1 or 2 but it can’t be 3 because r is smaller than 3.

So, the possible values of a are 3q or 3q + 1 or 3q + 2.

Now, find the square of all the possible values of a. If q is any positive integer, then its square (let’s call it “m”) will also be a positive integer.

Now, observe carefully that the square of all the positive integers is either of form 3m or 3m + 1 for some integer m.

We know that, a = 3q or 3q + 1 or 3q + 2

(a)2 = (3q)2 or (3q + 1)2 or (3q + 2)2

a2 = 3(3q2) or (9q2 + 6q + 1) or (9q2 + 12q + 4)

Case I: a2 = 3(3q2) where, m = 3q2

Case II: a2 = 3(3q2 + 2q) + 1 where, m = (3q2 + 2q)

Case III: a2 = 3(3q2 + 4q +1) + 1 where, m = (3q2 + 4q +1)

Thus, we see that a2 is of the form 3m or 3m + 1 where, m is any positive integer.

Hence, it can be said that the square of any positive integer is either of form 3m or 3m + 1.

Using Euclid's division lemma, it can be proved that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.