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Thread: Beal's conjecture conjecture

  1. #1 Beal's conjecture conjecture 
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    BEALíS CONJECTURE CONJECTURE
    Bealís Conjecture states that if A^x + B^y = C^z, x>2, y>2 and z>2 where A, B, C, x, y and z are positive integers then A, B and C must share a highest common factor. Using other variables one can restate Bealís conjecture i.e. if D^p + E^q = F^r, p>2, q>2 and r>2 where D, E, F, p, q and r are positive integers then D, E and F must share a highest common factor.
    Bealís Conjecture Conjecture
    All examples of Bealís Conjecture A^x + B^y = C^z (x>2, y>2 and z>2 where A, B, C, x, y and z are positive integers, x, y and z have different values can be manipulated to at least another example of Bealís Conjecture D^p + E^q = F^r (p>2, q>2 and r>2 where D, E, F, p, q and r are positive integers) where 2 of the 3 exponents p, q and r are equal where A^x = kD^p , B^y = kE^q , C^z = kF^r and k is a positive rational number constant that doesn't make p and/or q and/or or r less than or equal to 2.
    Example 7^6 + 7^7 = 98^3 can be manipulated to 49^3 + 7^7 = 98^3 where A=7, x=6, B=E=7, y=q=7, C=F=98, z=r=3, D=49, p=3 and k=1.
    Another example 242^10 + 242^11 = 175692^5 can be manipulated to 242^5 + 242^6 = 726^5 where A = 242, x=10, B=242, y=11, C=175692, z=5, D=242, p=5, E=242, q=6, F=726, r=5 and k= 1/242^5.
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  2. #2  
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    The point is to find one that is not that way or to prove the conjecture.
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