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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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The important question of class 10th maths of exercise 1.1 quesion no. 4 how can i solve easily because its very important question. Please help me for solving this 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.

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  1. Let x be any positive integer and y = 3.

    By Euclid’s division algorithm, then,

    x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

    Therefore, x = 3q, 3q+1 and 3q+2

    Now as per the question given, by squaring both the sides, we get,

    x2 = (3q)2 = 9q2 = 3 × 3q2

    Let 3q2 = m

    Therefore, x2= 3m ……………………..(1)

    x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1

    Substitute, 3q2+2q = m, to get,

    x2= 3m + 1 ……………………………. (2)

    x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1

    Again, substitute, 3q2+4q+1 = m, to get,

    x2= 3m + 1…………………………… (3)

    Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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