use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m Related: 02 - HCF & Euclid's Division Lemma - Class 10 - Maths?

Question

use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m Related:  02 - HCF & Euclid's Division Lemma - Class 10 - Maths?

Answer

Answer

the theorem starts like that : Consider natural numbers M, a, r. By Euclid's division lemma,

M = 3a + r where 0 ≤ r < 3

.

When r = 0, M = 3a.

M^2 mod 3 = (3a)^2 mod 3 =

0

.

So,

M^2 = 3m

.

When r = 1, M = 3a + 1.

M^2 mod 3 = (3a + 1)^2 mod 3.

As (3a + 1) mod 3 = 1,

M^2 mod 3 = 1^2 mod 3 =

1

.

So,

M^2 = 3m + 1

.

When r = 2, M = 3a + 2.

M^2 mod 3 = (3a + 2)^2 mod 3

= 2^2 mod 3 = 4 mod 3 =

1

.

So,

M^2 = 3m + 1

.

Thus, we've proved the theorem...

Answer

Using Euclid's Division Lemma to show the form of squares of positive integers
Euclid's Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < />

Proof:
- Let's consider a positive integer n.
- When n is divided by 3, there are only three possible remainders: 0, 1, or 2.
- Thus, n can be expressed as 3q, 3q + 1, or 3q + 2, where q is an integer.

Case 1: n = 3q
- The square of n is (3q)^2 = 9q^2 = 3(3q^2).
- This shows that the square of n is of the form 3m, where m = 3q^2.

Case 2: n = 3q + 1
- The square of n is (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1.
- This represents the square of n in the form 3m + 1, where m = 3q^2 + 2q.
Therefore, by using Euclid's Division Lemma and considering the three possible remainders when a positive integer is divided by 3, we have shown that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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