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Math Help - congruence modulo twenty four

  1. #1
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    congruence modulo twenty four

    1. if ab ≡ −1 (mod 24), then a + b ≡ 0 (mod 24).
    2. If n ≡ −1 (mod 24), then it is not a perfect square, and the number of divisors is even.
    3. If n ≡ −1 (mod 24), then 24 divides \sum_{d|n}d

    I don't need complete proofs, just some hints would be enough.
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  2. #2
    Super Member Bacterius's Avatar
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    Hello,
    (never mind what was there before)
    Choose any a. We have ab \equiv -1 \pmod{24}. Therefore, b \equiv -a^{-1} \pmod{24}. Plug this result back into the second condition : a + b \equiv a - a^{-1} \pmod{24}. Now you might want to take a look at what the multiplicative inverses are in \mathbb{Z}/24\mathbb{Z} (it is special)
    Last edited by Bacterius; March 22nd 2010 at 02:29 AM.
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  3. #3
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    Thanks, that proves 1, do you have a hint for 2 as well?
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  4. #4
    Super Member Bacterius's Avatar
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    Quote Originally Posted by EinStone View Post
    Thanks, that proves 1, do you have a hint for 2 as well?
    Sorry, I haven't had time to look at the other ones and it's bed time for me now
    Quickly looking, maybe proving that there are no perfect squares of the form 24k - 1 ? Maybe the prime factorization 24 = 2^3 \times 3 could help?
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  5. #5
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    Is this correct for 2.?

    Assume n ≡ −1 (mod 24) is a perfect square. Then n = c^2, thus c*c ≡ −1 (mod 24), by 1. we get c + c = 2c ≡ 0 (mod 24).

    so \exists M: 2c = M*24 => c= M*12 => c^2= 144M^2, hence n = c^2 \equiv 0\mod 24, a contradiction.

    And how to show that there is an even number of divisors?

    EDIT: Found 2.
    Need proof for 3.
    Last edited by EinStone; March 22nd 2010 at 03:20 PM. Reason: Solved partially
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