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Thread: Divisibility by 6

  1. #1
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    Divisibility by 6

    .


    Let a belong to the positive integers, and let p belong to the prime numbers, p > 3.


    Can you prove

    a(a + p)(a + 2p) is divisible by 6?
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  2. #2
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    Re: Divisibility by 6

    Let p be a prime with p>3, a a positive integer (actually any integer) and n=a(a+p)(a+2p). Then n is divisible by 6.

    First if a is even, then n is divisible by 2; if a is odd, a+p is even and again n is divisible by 2. So n is divisible by 2.

    Since p>3, p is not congruent to 0 mod 3. So p is 1 or 2 mod 3. Hence mod 3, n is congruent to a(a+1)(a+2) mod 3 or n is congruent to a(a+2)(a+1) mod 3. That is n is a(a+1)(a+2) mod 3. If a is 0, mod 3, 3 divides n. If a is 1 mod 3, a+2 is 0 mod 3 and again 3 divides n. If a is 2 mod 3, a+1 is 0 mod 3 and again 3 divides n.

    Since 2 divides n and 3 divides n, 6 divides n.
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  3. #3
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    Re: Divisibility by 6

    I would take a similar approach, but use the fact that p \equiv \pm 1 \pmod{6} as the starting point rather than p \equiv \pm 1 \pmod{3}. This probably removes all worries about a being odd or even although I haven't worked it through. It does add some complication to the evaluation of (a+p) \pmod{6} and (a+2p) \pmod{6} though.
    Last edited by Archie; Apr 24th 2017 at 08:42 AM.
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