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Math Help - [SOLVED] Proof with Primes

  1. #1
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    [SOLVED] Proof with Primes

    Prove that if p is a prime number and p is not equal to 3 then 3 divides p^2+2. (HINT:When p is divided by 3, the remainder is either 0, 1, or 2....i.e. for some k, p=3k or 3k+1 or 3k+2....)

    Ok, so for p to be prime, it has only 2 positive divisors (1 and itself).

    I tried doing this with a contradiction and assuuming 3 does not divide p^2+2, but I got a bit confused.

    I also tried something along the lines of Assume p is prime, then p=pk or p=1k for some integer k. But that just confused me. What can I do?
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    Prove that if p is a prime number and p is not equal to 3 then 3 divides p^2+2. (HINT:When p is divided by 3, the remainder is either 0, 1, or 2....i.e. for some k, p=3k or 3k+1 or 3k+2....)

    Ok, so for p to be prime, it has only 2 positive divisors (1 and itself).

    I tried doing this with a contradiction and assuuming 3 does not divide p^2+2, but I got a bit confused.

    I also tried something along the lines of Assume p is prime, then p=pk or p=1k for some integer k. But that just confused me. What can I do?
    It can be done with a direct computation. If P is a prime other than 3, then it is not divisible by 3 and so is of the form 3k+ 1 or 3k+ 2. If P= 3k+1 then P^2= 9k^2+ 6k+ 1 so P^2+ 2= 9k^2+ 6k+ 3= 3(3k^2+ 2k+ 1). If P= 3k+ 2 then P^2= 9k^2+ 12k+ 4 so P^2+ 2= 9k^2+ 12k+ 6= 3(3k^2+ 4k+ 2).
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  3. #3
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    Oh wow, thanks got it! That's the thing about proofs, sometimes they can be so obvious you don't see them!
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  4. #4
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    Where p > 3 then p^2 is at 6n+1, so your number is at 6n+3.
    Last edited by Whoever; February 10th 2009 at 10:41 AM.
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