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Math Help - A semiprime proof

  1. #1
    Super Member Bacterius's Avatar
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    A semiprime puzzle

    Hi all, here is a little conjecture (I do have my proof but it could be interesting to see others' proofs), let's see how you do.

    Say n = pq with p and q primes ( n is a semiprime, thus).

    Prove that there exists only two distinct solution pairs (a, m) such that m^2 - 4n = a^2 with m \in N, a \in N.

    Good luck !
    Last edited by Bacterius; November 16th 2009 at 08:12 PM.
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  2. #2
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    Are the solutions of (a,m) ( p-q , p+q ) , ( q-p ,p+q) ?
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  3. #3
    Super Member Bacterius's Avatar
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    Nearly : the solutions (a;m) are (n - 1;n + 1) and (q - p; p + q)
    Last edited by Bacterius; November 18th 2009 at 02:13 AM.
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  4. #4
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    i think simplependulum is right

    i got the same answer as simplependulum

    your first answer will get 2pq
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  5. #5
    Super Member Bacterius's Avatar
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    What to you mean by 2pq ?
    Anyway the point is to prove it, not give the two answers
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  6. #6
    Senior Member roninpro's Avatar
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    Hello. I noticed that you didn't actually receive a (formal) solution, so I thought that I would give mine.

    I will rearrange the equation in question:

     m^2-a^2=4n

    And once more:

     (m + a)(m - a) = 4pq

    We know for sure that the terms (m+a) and (m-a) are divisors of 4pq. Denoting d_1,d_2 as arbitrary divisors of 4pq, it suffices to solve all of the systems of the form m+a=d_1; m-a=d_2.

    The systems have integer solutions precisely when d_1=2pq,d_2=2 and d_1=2p,d_2=2q. In particular, m=pq+1,a=pq-1 and m=p+q,a=p-q, respectively. (Note that it may be necessary to reverse the roles of d_1,d_2 to adjust the sign of a.)

    Other combinations do not work for any semiprime or lead to redundancy when one of the factors is 2. For example, d_1=4,d_2=pq gives m=\frac{4+pq}{2},a=\frac{4-pq}{2}. We can see that a is always nonpositive, etc...

    I think that this does it, for the most part.

    Thanks for the problem.
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  7. #7
    Super Member Bacterius's Avatar
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    Interesting, indeed. My proof uses the fact that this equation can be considered as being the discriminant of the quadratic equation p^2 - mp + n = 0, where m = p + q. Therefore I can easily prove that there are only two solutions which are ... etc ...
    Good idea though, wouldn't have seen it like that
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