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Loser66

  • one year ago

How to prove a^3 -a divisible by 3 for all a? Please, help

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  1. SolomonZelman
    • one year ago
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    Factors into: *a(a-1)(a+1)*

  2. Loser66
    • one year ago
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    I know a^3 -a = a(a+1)(a-1) are 3 consecutive numbers, hence it divisible by3

  3. SolomonZelman
    • one year ago
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    so what's the prob/

  4. Loser66
    • one year ago
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    All I want is to go this way a is odd, a = 2k +1 a is even, a = 2k

  5. Loser66
    • one year ago
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    but I got stuck at a is odd.

  6. Loser66
    • one year ago
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    It is back to 3 consecutive numbers.

  7. SolomonZelman
    • one year ago
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    whether you start from odd or even integer, the product of 3 consequtive integers will be divisible by 3....

  8. Loser66
    • one year ago
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    :) Hence, no way to avoid "3 consecutive numbers" method?? right?OOOOOOk, that's all I want to know. Thank you.

  9. SolomonZelman
    • one year ago
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    Yes, 3 conseq.... (or not another way that I aware of using my small knowledge)

  10. freckles
    • one year ago
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    you could do induction if you don't like that method above... \[\text{ assume } 3i=a^3-a \text{ for some integer } a \\ ... \\ (a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1 \\ (a+1)^3-(a+1)=a^3-a+3a^2+3a \\ (a+1)^3-(a+1)=3i+3a^2+3a \\ (a+1)^3-(a+1)=3(i+a^2+a) \\ \text{ so } 3 \text{ is a factor of } (a+1)^3-(a+1)\]

  11. freckles
    • one year ago
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    and i is an integer of course

  12. Loser66
    • one year ago
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    Got you. Thanks @freckles

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