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Math Help - I have problem about primes.(Help me please!)

  1. #1
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    Post I have problem about primes.(Help me please!)

    If p>=q>=5 and p and q are both primes, prove that 24|(p^2-q^2).



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  2. #2
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    Quote Originally Posted by konna View Post
    If p>=q>=5 and p and q are both primes, prove that 24|(p^2-q^2).



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    What have you tried yourself? Of course, p^2- q^2= (p- q)(p+q) since p and q are both primes larger than 3, they are both odd and so both p-q and p+q are even- their product is, at least, divisible by 4. Now see if you can find another factor of 2. If p= 2m+1 and q= 2n+1, p+q= 2m+2n+ 2= 2(m+n+1) and p- q= 2m- 2n= 2(m-n). Consider what happens to m+n+1 and m-n if m and n are both even, both odd, or one even and the other odd. Notice that so far we have only required that p and q be odd, not that they be prime.

    Once you have done that the only thing remaining is to show that p^2- q^2 must be a multiple of 3 and that means showing that either p-q or p+q is a multiple of 3.
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  3. #3
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    Quote Originally Posted by konna View Post
    If p>=q>=5 and p and q are both primes, prove that 24|(p^2-q^2).



    Thank.

    p^2-q^2=(p-q)(p+q) , and as:

    1) x^2=1\!\!\!\!\pmod 3\,\,\,\forall\,x not a multiple of 3 ;

    2) either p-q or p+q is divisible by 4, and the other one by 2.

    The result follows at once.

    Tonio
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  4. #4
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    I used to try by myself , but i not sure my process.

    Thank so much ..
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  5. #5
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    Hello, konna!

    I don't know if this qualifies as a proof.


    If p \geq q \geq 5 and p and q are both primes,

    . . prove that: . 24\,|\,(p^2-q^2)

    Any prime greater than or equal to 5 is of the form: . 6m \pm 1
    . . for some integer m.


    Let: . \begin{array}{ccc}p &=& 6m \pm 1 \\ q &=& 6n \pm1\end{array}

    Then: . \begin{array}{ccc}p^2 &=& 36m^2 \pm 12m + 1 \\ q^2 &=& 36n^2 \pm12n + 1 \end{array}

    Subtract: . p^2-q^2 \;=\;36m^2-36n^2 \pm 12m \mp 12n \;=\;36(m^2-n^2) \pm12(m - n)

    . . . . . . . p^2-q^2 \;=\;36(m-n)(m+n) \pm12(m - n)
    . . . . . . . p^2-q^2 \;=\;12(m-n)\bigg[3(m+n) \pm 1\bigg]


    We see that p^2-q^2 is divisible by 12.

    We must show that either (m-n) or [3(m+n) \pm1] is even.


    If m and n has the same parity (both even or both odd),
    . . then (m-n) is even.


    If m and n have opposite parity (one even, one odd),
    . . then (m+n) is odd.
    And . 3(m+n) is odd.
    . . Then: . 3(m+n) \pm1 is even.


    Therefore, p^2-q^2 is divisible by 24.

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