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.

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.

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2 thoughts on “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.”

  1. Step-by-step explanation:

    Sol: Let a be a positive integer q be the quotient and r be the remainder

    Dividing a by 3 using the Euclid’s Division Lemma

    we have

    a = 3q + r,

    where a = ≤ r < 3

    Putting r = 0, 1 and 2, we get:

    a = 3q

    → a^2 = 9q^2

    = 3 × 3q^2

    = 3m (Assuming m = q^2)

    Then, a = 3q + 1

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

    = 9q^2 + 6q + 1

    = 3(3q^2 + 2q) + 1

    = 3m +1 (Assuming m = 3q^2 + 2q)

    Next, a = 3q + 2

    → a² = (3q + 2)^2

    = 9q^2 + 12q + 4

    = 9q^2 + 12q +3+1

    = 3(3q^2 + 4q + 1) + 1

    = 3m + 1. (Assuming m = 3q^2 + 4q+1)

    Therefore, the square of any positive integer (say a^2) is always of the form 3m or 3m + 1

    Hence, proved

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