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urbanderivative

  • one year ago

prove that the square of any integer a is either of the form 3k or 3k+1 for some integer k. [hint by division algorithm, a must be of the form 2q, 3q +1, or 3q+2]

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  1. matricked
    • one year ago
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    any integer can be written as 3q, 3q +1, or 3q+2....q being an integer case 1 when n=3q clearly n^2=9q^2=3(3q^2)

  2. matricked
    • one year ago
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    when n=3q +1 n^2=9q^2+6q+1=3(3q^2+2q)+1

  3. matricked
    • one year ago
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    when n=3q +2 n^2=9q^2+12q+4=3(3q^2+4q+1)+1 (form 3k+1)

  4. matricked
    • one year ago
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    the square of any integer a is either of the form 3k or 3k+1

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