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JoãoVitorMC

Prove or disprove: For any integer n, the number n² + n + 1 is odd.

  • one year ago
  • one year ago

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  1. AriPotta
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    i think that's true

    • one year ago
  2. JoãoVitorMC
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    i think that to but i have to prove some how

    • one year ago
  3. AriPotta
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    5^2 + 5 +1 25 + 5 +1 = 31 (odd) 1^2 + 1 + 1 1 + 1 +1 = 3 (odd)

    • one year ago
  4. AriPotta
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    so we just proved it

    • one year ago
  5. jim_thompson5910
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    break it up into cases Case 1: If n is even,... Case 2: If n is odd,...

    • one year ago
  6. AriPotta
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    i can't find any number that disproves it...

    • one year ago
  7. jim_thompson5910
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    AriPotta you proved it works for those particular values, but you didn't prove it for all integers

    • one year ago
  8. AriPotta
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    2^2 + 2 + 1 4 + 2 + 1 = 7 (odd)

    • one year ago
  9. JoãoVitorMC
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    ok i'll try using cases

    • one year ago
  10. jim_thompson5910
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    Case 1: n is an even integer Let k be any integer, so 2k is always even. Let n = 2k n^2 + n + 1 (2k)^2 + (2k) + 1 4k^2 + 2k + 1 2(2k^2 + k) + 1 2q + 1 ... Let q = 2k^2 + k (you can easily show that q is always an integer) So if n = 2k (ie if n is an even integer), then n^2 + n + 1 is equivalent to 2q + 1, which is an odd integer. I'll let you do case 2

    • one year ago
  11. JoãoVitorMC
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    thank you!

    • one year ago
  12. benzilla
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    An even + an odd is an odd . an odd + odd is an even number, and an even + even is an even number. If n is odd n^2 is odd so we have an odd + odd + 1 which is the same as odd + even so it will be odd. If n is an even n^2 is even so n^2 +n will be even. If you add 1 to an even number it is odd. Aripotta has the right idea. Testing numbers works for odd and even in all cases.

    • one year ago
  13. jim_thompson5910
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    np

    • one year ago
  14. JoãoVitorMC
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    thank you all of you!

    • one year ago
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