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Math Help - Proof of a Conjecture or Counter Examples

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    Proof of a Conjecture or Counter Examples

    {P-Q}{P^(n-1) + K1[P^(n-2)]Q + K2[P^(n-3)]Q^2 + ....+ Q^(n-1)} cannot equal C^n - B^n
    where all the variables are positive integers, n>2, P>Q, C>B
    unless all the Ks are equal to 1 or in some instances the Ks have the same value.
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    Re: Proof of a Conjecture or Counter Examples

    Can anyone please prove or disprove this?
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    Re: Proof of a Conjecture or Counter Examples

    Please can I have another counterexample, if any, because this one kind off eliminates Q by setting it equal to 1. So an additional constraint of Q > 1.
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    Re: Proof of a Conjecture or Counter Examples

    Please don't bump your threads.

    -Dan
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    Re: Proof of a Conjecture or Counter Examples

    Quote Originally Posted by topsquark View Post
    Please don't bump your threads.

    -Dan
    Can i ask why my post was deleted? It was a completely legitimate post providing a proper counterexample.

    PS: I have another counterexample with the new constraint given, but if someone is deleting links for no reason, there is no reason to post it.

    EDIT:
    Forget my above comments. It is just the TS created 2 exact topics and i answered to the other:
    Proof of a Conjecture or Counter Examples
    Last edited by ChessTal; July 23rd 2013 at 11:46 AM.
    Thanks from topsquark
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    Re: Proof of a Conjecture or Counter Examples

    So what's your other counterexample please?
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    Forum Admin topsquark's Avatar
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    Re: Proof of a Conjecture or Counter Examples

    First, take a look at the example in the other thread and see if you can come up with another one based on ChessTal's method.

    This is why we don't double post....

    -Dan
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    Re: Proof of a Conjecture or Counter Examples

    Quote Originally Posted by MrAwojobi View Post
    So what's your other counterexample please?
    For:
    P=3
    Q=2
    n=3
    K1=83
    C=8
    B=1 we have:

    (P - Q) \cdot \left( {{P^{n - 1}} + {K_1} \cdot {P^{n - 2}} \cdot  {Q^1} + {K_2} \cdot {P^{n - 3}} \cdot {Q^2} + ... + {K_{n - 2}} \cdot  {P^1} \cdot {Q^{n - 2}} + {K_{n - 1}} \cdot {Q^{n - 1}}} \right) \ne  {C^n} - {B^n} \Leftrightarrow

    (3 - 2) \cdot ({3^{3 - 1}} + 83 \cdot {3^{3 - 2}} \cdot {2^1} + {2^{3 - 1}}) \ne {8^3} - {1^3} \Leftrightarrow

    9 + 498 + 4 \ne 511 \Leftrightarrow

    511 \ne 511
    Last edited by ChessTal; July 23rd 2013 at 12:02 PM.
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    Re: Proof of a Conjecture or Counter Examples

    Many thanks for these counterexamples. However can I impose one last constraint i.e. (P-Q) > 1
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    Re: Proof of a Conjecture or Counter Examples

    Quote Originally Posted by MrAwojobi View Post
    Many thanks for these counterexamples. However can I impose one last constraint i.e. (P-Q) > 1
    Hmm i guess it would be easy to find another counterexample but we will see.
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    Re: Proof of a Conjecture or Counter Examples

    Thanks.
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    Re: Proof of a Conjecture or Counter Examples

    Well another one is:
    P=4
    Q=2
    n=3
    K1=43
    C=9
    B=1
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    Re: Proof of a Conjecture or Counter Examples

    Thank you again. Please don't get tired of me yet but I promise this is the last constraint. P and Q should not share any common factors except the trivial unity.
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    Re: Proof of a Conjecture or Counter Examples

    Quote Originally Posted by MrAwojobi View Post
    Thank you again. Please don't get tired of me yet but I promise this is the last constraint. P and Q should not share any common factors except the trivial unity.
    Damn. If a conjecture is wrong, is wrong!!

    With the new constraint(LCD(5,2)=1) the conjecture is still wrong:
    P=5
    Q=2
    n=3
    K1=352
    C=22
    B=1
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    Re: Proof of a Conjecture or Counter Examples

    B>1, C and B don't share a common factor except the trivial unity are the last set of constraints. Sorry I broke my promise.
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