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Math Help - Prove that ab + cd is not prime.

  1. #1
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    Prove that ab + cd is not prime.

    Let a, b, c, d
    be integers with a > b > c > d > 0.
    Suppose that
    ac + bd = (b + d + a − c)(b + d − a + c).
    Prove that ab + cd is not prime.
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  2. #2
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    Quote Originally Posted by perash View Post
    Let a, b, c, d
    be integers with a > b > c > d > 0.
    Suppose that
    ac + bd = (b + d + a − c)(b + d − a + c).
    Prove that ab + cd is not prime.
    b+d+a-c>0, so either ac+bd=0 and b+d-a+c=0, or ac+db>0 and b+d-a+c>0

    So we have either ac+bd=0, or there exist positive integers u, v such that ac+db=uv. In either case ac+db is not prime.

    RonL
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  3. #3
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    In either case ac+db is not prime.
    How does that show that \color{red}ab+cd is not prime?
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    answer

    if ac + bd = (b+d+a-c)(b+d-a+c)
    we can factorise it, because (b+d+a-c)(b+d-a+c) is a difference of perfect squares
    (b+d)^2-(a-c)^2
    expands to b^2+d^2-a^2-c^2+2bd+2ac
    factor of 2 here
    b^2+d^2-a^2-c^2+2(a+b+c+d)
    because there is a factor of two, the number is even and therefore cannot be prime
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  5. #5
    Senior Member JaneBennet's Avatar
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    Look. Are some people blind or what? The question is asking to show that ab+cd NOT ac+bd is not prime.

    Quote Originally Posted by perash View Post
    Let a, b, c, d
    be integers with a > b > c > d > 0.
    Suppose that
    ac + bd = (b + d + a − c)(b + d − a + c).
    Prove that ab + cd is not prime.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by JaneBennet View Post
    How does that show that \color{red}ab+cd is not prime?
    It doesn't, because I misread the question

    RonL
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JaneBennet View Post
    Look. Are some people blind or what? The question is asking to show that ab+cd NOT ac+bd is not prime.
    Wow, you're mean.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Wow, you're mean.
    She's not mean she's speaking her mind. I have a great deal of respect for those who do so.

    -Dan
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    She's not mean she's speaking her mind. I have a great deal of respect for those who do so.

    -Dan
    You are right, she is just an individual speaking her mind!....a mean individual
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    [snip]....a mean individual
    I think JB is a bit more than average ......

    (*sigh* a desperate attempt at a statistics joke ...... )
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  11. #11
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    mrs mean

    well if speaking ur mind is so acceptable, u r mean. in fact ur terrible. the question is to prove it true, but it does prove it. there is a factor of two in the entire thing. have a god day
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    Quote Originally Posted by abes22 View Post
    well if speaking ur mind is so acceptable, u r mean. in fact ur terrible. the question is to prove it true, but it does prove it. there is a factor of two in the entire thing. have a god day
    The question is given

    a, b, c, d integers with a > b > c > d > 0, and:


    ac + bd = (b + d + a − c)(b + d − a + c).

    Prove that ab + cd is not prime.

    What you have proven is that ac + bd is not prime not that ab + cd is not prime. Which was the whole point of Jane's first post.

    It is an easy mistake to make, I know because I made it myself.

    (Also your argument could do with some work it is not entirly clear what you are trying to do. Also maybe your language could do with some moderation as well it is close to getting an infraction for being insulting)

    RonL
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  13. #13
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    you're right, although then again I'm 15.
    But that lady whoever she may be had no right to say something like that. Regardless of whether or not i did misread the question. So my insult was equally as bad, and I wasn't prepared to let her go unnoticed.
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  14. #14
    Forum Admin topsquark's Avatar
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    I think enough has been said about this. Please only add a post here if you can contribute to the problem that has been asked.

    -Dan
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    a x^2+b x+c=0
    x=\frac{-b\pm \sqrt{b^2-4 a c}}{2 a}

    Therefore if (a,b,c,x) are integers, than x is divided by 2. [1]


    You have
    a c+b d=(a+b-c+d) (-a+b+c+d)
     a c+b d=a^2-b^2+c^2-d^2

    This reduces to a quadratic equation for a: a^2-\text{ac}-b^2+c^2+b d-d^2=0
    So a is an even number (from 1). The same can be proved for any of the (a,b,c,d).
    Because of that a*b+c*d=even*even+even*even=even+even=even
    Even numbers are not primes except for 2. But the smallest values of a,b,c,d are:
    d=1
    c=2
    b=3
    a=4
    So:
    \text{ab}+\text{cd}=4*3+2*1\neq 2=>2 CAN'T be a solution
    =>ab+cd=even \neq2=>ab+cd is not a prime

    Hope that helps
    Last edited by fobos3; June 30th 2008 at 07:46 AM.
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