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Thread: [SOLVED] prime number problem

  1. #1
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    [SOLVED] prime number problem

    I need a hel for find out the answers for the followinf number theory, prime number questions, please give me a help to find out the answers

    1.) If p is prime prove that there exist no positive integers a and b s.t. a^2 = pb^2

    2.) If [(2^n) -1] is a prime, the n is also a prime

    3.) If n>= 5 is a prime number, show that n^2 + 2 is not a prime.

    4.) If b(not= 2) is a prime number show that b^2 + (b + 2)^2 + (b + 4)^2 +1 is divisible by 12.

    please see the attached file for better understanding of the question
    Attached Thumbnails Attached Thumbnails [SOLVED] prime number problem-nq1.jpg  
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    1.) $\displaystyle a^2=pb^2 \Longrightarrow p=\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2 $.

    Hence we've shown that $\displaystyle p=c^2 $ so $\displaystyle c \mid p $ where $\displaystyle 1<c<p $. This means $\displaystyle p $ is not prime. So we've reached a contradiction.

    (Note $\displaystyle c \neq 1 $ because otherwise this would imply that $\displaystyle p = 1 $ which we know to be false.)
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    2.) It's easiest to prove the contrapositive:
    If $\displaystyle n $ is not prime then $\displaystyle 2^n-1 $ is not prime.

    Since $\displaystyle n $ is composite we know that $\displaystyle n=ab $ where $\displaystyle 1 < a,b < n $.

    Now let's consider $\displaystyle 2^n-1 $.

    $\displaystyle 2^n-1 = 2^{ab}-1=(2^a-1)\left(1+2^a+2^{2a}+2^{3a}+\cdots+2^{(b-1)a}\right) $

    Hence $\displaystyle 2^n-1 $ is also composite.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    4.) $\displaystyle b^2+(b+2)^2+(b+4)^2+1 = 3(b^2+4b+7) $
    Therefore we must show $\displaystyle b^2+4b+7 $ is divisible by $\displaystyle 4 $.

    This is easy to see since we know $\displaystyle b $ is odd, so $\displaystyle b \equiv \pm 1 \mod{4} \Longrightarrow b^2 \equiv 1 \mod{4} $.

    So we have $\displaystyle b^2+4b+7 \equiv 1+0+7 \equiv 8 \equiv 0 \mod{4} $. Hence $\displaystyle b^2+4b+7 $ is divisible by $\displaystyle 4 $.

    (Notice we didn't use the fact that $\displaystyle b $ was prime. This result holds for all odd integers.)
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
    4.) $\displaystyle b^2+(b+2)^2+(b+4)^2+1 = 3(b^2+4b+7) $
    Therefore we must show $\displaystyle b^2+4b+7 $ is divisible by $\displaystyle 4 $.

    This is easy to see since we know $\displaystyle b $ is odd, so $\displaystyle b \equiv \pm 1 \mod{4} \Longrightarrow b^2 \equiv 1 \mod{4} $.

    So we have $\displaystyle b^2+4b+7 \equiv 1+0+7 \equiv 8 \equiv 0 \mod{4} $. Hence $\displaystyle b^2+4b+7 $ is divisible by $\displaystyle 4 $.

    (Notice we didn't use the fact that $\displaystyle b $ was prime. This result holds for all odd integers.)

    3. Hint: What can you say about $\displaystyle 3|(n^2+2)$?
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  6. #6
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    3: Every prime number $\displaystyle \geq 5$ can be expressed as $\displaystyle p=6k \pm 1$ for a natural number k. Make sure that you understand why this is true and use this fact together with the hint from aman_cc to solve the problem.
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