REAL NUMBERS | Irrational Numbers | Ex -1

Using Euclid's Division Lemma to Determine the Square Form of Any Positive Integer

Question:

Use Euclid's Division Lemma to show that the square of any positive integer is either of the form \( 3m \) or \( 3m + 1 \), where \( m \) is an integer.

Solution:

According to Euclid's Division Lemma, for any integer \( n \) and a fixed positive integer divisor \( d = 3 \), there are unique integers \( q \) and \( r \) such that:

\( n = 3q + r \quad \,where \, r = 0, 1,\, or \, 2 \)

We now consider each case:

Case 1: \( r = 0 \)

\[ n = 3q \Rightarrow n^2 = (3q)^2 = 9q^2 = 3(3q^2) \] Hence, \( n^2 \) is a multiple of 3 and can be expressed in the form \( 3m \).

Case 2: \( r = 1 \)

\[ n = 3q + 1 \Rightarrow n^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 \] So, \( n^2 \) is of the form \( 3m + 1 \).

Case 3: \( r = 2 \)

\[ n = 3q + 2 \Rightarrow n^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 \] So, again, \( n^2 \) is of the form \( 3m + 1 \).

Therefore, for any positive integer \( n \), the square \( n^2 \) is always either of the form:

• \( 3m \) — when \( n \equiv 0 \mod 3 \)
• \( 3m + 1 \) — when \( n \equiv 1 \mod 3 \) or \( n \equiv 2 \mod 3 \)

Hence, the square of any positive integer is always of the form \( 3m \) or \( 3m + 1 \), as required.