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gmer Group Title

Prove that if n is an integer, then n^2 mod 5 is either 0,1, or 4

  • one year ago
  • one year ago

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  1. Shadowys Group Title
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    there are only five possible ways for n. first case: \(n \equiv 0(mod 5)\), then \(n^2 \equiv 0(mod 5)\) second case: \(n \equiv 1(mod 5)\)), then \(n^2 \equiv 1(mod 5)\) third case:\(n \equiv 2(mod 5)\)), then \(n^2 \equiv 4(mod 5)\) fourth case:\(n \equiv 3(mod 5)\)), then \(n^2 \equiv 6(mod 5)\), which is equal to \(n^2 \equiv 1(mod 5)\), as \(6 \equiv 1(mod 5)\) fifth case:\(n \equiv 4(mod 5)\)), then \(n^2 \equiv 16(mod 5)\) which is equal to \(n^2 \equiv 1(mod 5)\), as \(16 \equiv 1(mod 5)\) so we conclude that if n is an integer, n^2 mod 5 is either 0,1, or 4

    • one year ago
  2. Shadowys Group Title
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    sorry, for fourth case, it's \(n^2 \equiv 9 (mod 5)\) and it's equal to \(n^2 \equiv 4 (mod 5)\)

    • one year ago
  3. gmer Group Title
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    Excellent, but how do we know that there are only five possible values for n?

    • one year ago
  4. Shadowys Group Title
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    because the remainder for any integer k divided by an integer n has only n-1 remainders. e.g. for 2, you can only get 0 or 1 as remainder. any other remainder means that the division is incomplete.

    • one year ago
  5. Shadowys Group Title
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    sorry it'snot n-1, it's n. but the range is 0,1,2,3,...,n-1

    • one year ago
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