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September 28th, 2015, 10:58 AM  #1 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4  Simple Proof of Fermat's Last Theorem and Beal's Conjecture
Fermat’s Last Theorem The theorem states that Aⁿ + Bⁿ ≠ Cⁿ if n > 2, A, B and C are all positive integers. Andrew Wiles produced a lengthy proof of over 100 pages in the mid 1990s. If A, B and C have a highest common factor f > 1, fⁿ can be cancelled out to give a new Fermat inequation with new values for A, B and C. In order to prove the theorem it is sufficient therefore to consider only cases where A, B and C have a highest common factor f = 1 i.e. ‘primitive’ inequations. Beal’s Conjecture: If P^x + Q^y = R^z where P, Q, R, x, y and z are positive integers, x > 2, y > 2 and z > 2 then P, Q and R must have a highest common factor > 1. Proof of Fermat’s Last Theorem Fermat’s inequation can be rewritten as Cⁿ  Bⁿ ≠ Aⁿ where A, B and C have a highest common factor f = 1. Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)/F = I^n C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = F I^n  B^(n1)  C^(n1) C B[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3) = FI^n  B^(n1)  C^(n1) This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ, hence the proof. Proof of Beal’s Conjecture Let it be initially assumed that P, Q and R have a highest common factor = 1. P^x + Q^y = R^z can be rewritten as {R^(z/n)  Q^(y/n)}([R^(z/n)]^(n1)+ [R^(z/n)]^(n2) [Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)} = P^x. When n is odd an alternative equation is [P^(x/n)]^n + [Q^(y/n)]^n = R^z {P^(x/n) + Q^(y/n) [[P^(x/n)]^(n1) [P^(x/n)]^(n2) [Q^(y/n)] + …..  [P^(x/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)= R^z where n > 2 is a positive integer. For the equation written in these factorized forms to produce integer values for P^x and R^z, it should be clear that P^(x/n) , Q^(y/n) and R^(z/n) must be integers. It is no wonder that all examples of Beal’s equation can be derived from equations of either of these forms in which two exponents are the same. Let P^(x/n) = A, Q^(y/n) = B and R^(z/n) = C where A, B and C have a highest common factor = 1 since P, Q and R have a highest common factor = 1. {R^(z/n)  Q^(y/n)} {[R^(z/n)]^(n1)+ [R^(z/n)]^(n2)[Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) +[Q^(y/n)]^(n1)} = [C  B][C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1). Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)/F] Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2(C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider (Cⁿ1 + Cn2B + ….. + CBⁿ2 + Bⁿ1)/F = I^n Cⁿ1 + Cn2B + ….. + CBⁿ2 + Bⁿ1 = FI^n C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = F I^n  B^(n1)  C^(n1) C B[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3) = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ. The only way for FI^n  B^(n1)  C^(n1) to simultaneously be a multiple of B and C is if B, C and therefore A have a highest common factor > 1. This proves the validity of Beal’s conjecture. Fermat’s Last Theorem: The theorem states that Aⁿ + Bⁿ ≠ Cⁿ if n > 2, A, B and C are all positive integers. Andrew Wiles produced a lengthy proof of over 100 pages in the mid 1990s. If A, B and C have a highest common factor f > 1, fⁿ can be cancelled out to give a new Fermat inequation with new values for A, B and C. In order to prove the theorem it is sufficient therefore to consider only cases where A, B and C have a highest common factor f = 1 i.e. ‘primitive’ inequations. Beal’s Conjecture: If P^x + Q^y = R^z where P, Q, R, x, y and z are positive integers, x>2, y>2 and z>2 then P, Q and R must have a highest common factor > 1. Proof of Fermat’s Last Theorem: Fermat’s inequation can be rewritten as Cⁿ  Bⁿ ≠ Aⁿ where A, B and C have a highest common factor f = 1. Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ, hence the proof. Proof of Beal’s Conjecture: Let it be initially assumed that P, Q and R have a highest common factor = 1. P^x + Q^y = R^z can be rewritten as {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2) [Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)} = P^x. When n is odd an alternative equation is [P^(x/n)]^n + [Q^(y/n)]^n = R^z. [P^(x/n) + Q^(y/n)][P^(x/n)]^(n1)  [P^(x/n)]^(n2) [Q^(y/n)] + …..  [P^(x/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)= R^z where n > 2 is a positive integer. For the equation written in these factorized forms to produce integer values for P^x and R^z, it should be clear that P^(x/n) , Q^(y/n) and R^(z/n) must be integers. It is no wonder that all examples of Beal’s equation can be derived from equations of either of these forms in which two exponents are the same. Let P^(x/n) = A, Q^(y/n) = B and R^(z/n) = C where A, B and C have a highest common factor = 1 since P, Q and R have a highest common factor = 1. {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2)[Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) +[Q^(y/n)]^(n1)} = [C  B][C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. The only way for FI^n  B^(n1)  C^(n1) to simultaneously be a multiple of B and C is if B, C and therefore A have a highest common factor > 1. This proves the validity of Beal’s conjecture. Fermat’s Last Theorem: The theorem states that Aⁿ + Bⁿ ≠ Cⁿ if n > 2, A, B and C are all positive integers. Andrew Wiles produced a lengthy proof of over 100 pages in the mid 1990s. If A, B and C have a highest common factor f > 1, fⁿ can be cancelled out to give a new Fermat inequation with new values for A, B and C. In order to prove the theorem it is sufficient therefore to consider only cases where A, B and C have a highest common factor f = 1 i.e. ‘primitive’ inequations. Beal’s Conjecture: If P^x + Q^y = R^z where P, Q, R, x, y and z are positive integers, x>2, y>2 and z>2 then P, Q and R must have a highest common factor > 1. Proof of Fermat’s Last Theorem: Fermat’s inequation can be rewritten as Cⁿ  Bⁿ ≠ Aⁿ where A, B and C have a highest common factor f = 1. Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ, hence the proof. Proof of Beal’s Conjecture: Let it be initially assumed that P, Q and R have a highest common factor = 1. P^x + Q^y = R^z can be rewritten as {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2) [Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)} = P^x. When n is odd an alternative equation is [P^(x/n)]^n + [Q^(y/n)]^n = R^z. [P^(x/n) + Q^(y/n)][P^(x/n)]^(n1)  [P^(x/n)]^(n2) [Q^(y/n)] + …..  [P^(x/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)= R^z where n > 2 is a positive integer. For the equation written in these factorized forms to produce integer values for P^x and R^z, it should be clear that P^(x/n) , Q^(y/n) and R^(z/n) must be integers. It is no wonder that all examples of Beal’s equation can be derived from equations of either of these forms in which two exponents are the same. Let P^(x/n) = A, Q^(y/n) = B and R^(z/n) = C where A, B and C have a highest common factor = 1 since P, Q and R have a highest common factor = 1. {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2)[Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) +[Q^(y/n)]^(n1)} = [C  B][C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. The only way for FI^n  B^(n1)  C^(n1) to simultaneously be a multiple of B and C is if B, C and therefore A have a highest common factor > 1. This proves the validity of Beal’s conjecture. Fermat’s Last Theorem: The theorem states that Aⁿ + Bⁿ ≠ Cⁿ if n > 2, A, B and C are all positive integers. Andrew Wiles produced a lengthy proof of over 100 pages in the mid 1990s. If A, B and C have a highest common factor f > 1, fⁿ can be cancelled out to give a new Fermat inequation with new values for A, B and C. In order to prove the theorem it is sufficient therefore to consider only cases where A, B and C have a highest common factor f = 1 i.e. ‘primitive’ inequations. Beal’s Conjecture: If P^x + Q^y = R^z where P, Q, R, x, y and z are positive integers, x>2, y>2 and z>2 then P, Q and R must have a highest common factor > 1. Proof of Fermat’s Last Theorem: Fermat’s inequation can be rewritten as Cⁿ  Bⁿ ≠ Aⁿ where A, B and C have a highest common factor f = 1. Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ, hence the proof. Proof of Beal’s Conjecture: Let it be initially assumed that P, Q and R have a highest common factor = 1. P^x + Q^y = R^z can be rewritten as {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2) [Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)} = P^x. When n is odd an alternative equation is [P^(x/n)]^n + [Q^(y/n)]^n = R^z. [P^(x/n) + Q^(y/n)][P^(x/n)]^(n1)  [P^(x/n)]^(n2) [Q^(y/n)] + …..  [P^(x/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)= R^z where n > 2 is a positive integer. For the equation written in these factorized forms to produce integer values for P^x and R^z, it should be clear that P^(x/n) , Q^(y/n) and R^(z/n) must be integers. It is no wonder that all examples of Beal’s equation can be derived from equations of either of these forms in which two exponents are the same. Let P^(x/n) = A, Q^(y/n) = B and R^(z/n) = C where A, B and C have a highest common factor = 1 since P, Q and R have a highest common factor = 1. {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2)[Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) +[Q^(y/n)]^(n1)} = [C  B][C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. The only way for FI^n  B^(n1)  C^(n1) to simultaneously be a multiple of B and C is if B, C and therefore A have a highest common factor > 1. This proves the validity of Beal’s conjecture. Fermat’s Last Theorem: The theorem states that Aⁿ + Bⁿ ≠ Cⁿ if n > 2, A, B and C are all positive integers. Andrew Wiles produced a lengthy proof of over 100 pages in the mid 1990s. If A, B and C have a highest common factor f > 1, fⁿ can be cancelled out to give a new Fermat inequation with new values for A, B and C. In order to prove the theorem it is sufficient therefore to consider only cases where A, B and C have a highest common factor f = 1 i.e. ‘primitive’ inequations. Beal’s Conjecture: If P^x + Q^y = R^z where P, Q, R, x, y and z are positive integers, x>2, y>2 and z>2 then P, Q and R must have a highest common factor > 1. Proof of Fermat’s Last Theorem: Fermat’s inequation can be rewritten as Cⁿ  Bⁿ ≠ Aⁿ where A, B and C have a highest common factor f = 1. Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ, hence the proof. Proof of Beal’s Conjecture: Let it be initially assumed that P, Q and R have a highest common factor = 1. P^x + Q^y = R^z can be rewritten as {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2) [Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)} = P^x. When n is odd an alternative equation is [P^(x/n)]^n + [Q^(y/n)]^n = R^z. [P^(x/n) + Q^(y/n)][P^(x/n)]^(n1)  [P^(x/n)]^(n2) [Q^(y/n)] + …..  [P^(x/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)= R^z where n > 2 is a positive integer. For the equation written in these factorized forms to produce integer values for P^x and R^z, it should be clear that P^(x/n) , Q^(y/n) and R^(z/n) must be integers. It is no wonder that all examples of Beal’s equation can be derived from equations of either of these forms in which two exponents are the same. Let P^(x/n) = A, Q^(y/n) = B and R^(z/n) = C where A, B and C have a highest common factor = 1 since P, Q and R have a highest common factor = 1. {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2)[Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) +[Q^(y/n)]^(n1)} = [C  B][C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. The only way for FI^n  B^(n1)  C^(n1) to simultaneously be a multiple of B and C is if B, C and therefore A have a highest common factor > 1. This proves the validity of Beal’s conjecture. Last edited by CRGreathouse; September 29th, 2015 at 08:39 AM. 
September 29th, 2015, 08:22 AM  #2 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
Fermat’s Last Theorem: The theorem states that Aⁿ + Bⁿ ≠ Cⁿ if n > 2, A, B and C are all positive integers. Andrew Wiles produced a lengthy proof of over 100 pages in the mid 1990s. If A, B and C have a highest common factor f > 1, fⁿ can be cancelled out to give a new Fermat inequation with new values for A, B and C. In order to prove the theorem it is sufficient therefore to consider only cases where A, B and C have a highest common factor f = 1 i.e. ‘primitive’ inequations. Beal’s Conjecture: If P^x + Q^y = R^z where P, Q, R, x, y and z are positive integers, x>2, y>2 and z>2 then P, Q and R must have a highest common factor > 1. Proof of Fermat’s Last Theorem: Fermat’s inequation can be rewritten as Cⁿ  Bⁿ ≠ Aⁿ where A, B and C have a highest common factor f = 1. Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. This means that Cⁿ  Bⁿ ≠ Aⁿ, hence the proof. Proof of Beal’s Conjecture: Let it be initially assumed that P, Q and R have a highest common factor = 1. P^x + Q^y = R^z can be rewritten as {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2) [Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)} = P^x. When n is odd an alternative equation is [P^(x/n)]^n + [Q^(y/n)]^n = R^z. [P^(x/n) + Q^(y/n)][P^(x/n)]^(n1)  [P^(x/n)]^(n2) [Q^(y/n)] + …..  [P^(x/n)] [Q^(y/n)]^(n2) + [Q^(y/n)]^(n1)= R^z where n > 2 is a positive integer. For the equation written in these factorized forms to produce integer values for P^x and R^z, it should be clear that P^(x/n) , Q^(y/n) and R^(z/n) must be integers. It is no wonder that all examples of Beal’s equation can be derived from equations of either of these forms in which two exponents are the same. Let P^(x/n) = A, Q^(y/n) = B and R^(z/n) = C where A, B and C have a highest common factor = 1 since P, Q and R have a highest common factor = 1. {R^(z/n)  Q^(y/n)}{[R^(z/n)]^(n1)+ [R^(z/n)]^(n2)[Q^(y/n)] + ….. + [R^(z/n)] [Q^(y/n)]^(n2) +[Q^(y/n)]^(n1)} = [C  B][C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1)]. Let F be the highest common factor of CB and C^(n1) + C^(n2) B + ….. + CB^(n2) + B^(n1). Therefore Cⁿ  Bⁿ = [C  B][C^(n1) + C^(n2)B + ….. + C B^(n2) + B^(n1)] = F(C  B)/F[F{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = F^2 (C  B)/F[{C^(n1) + C^(n2)B + ….. + CB^(n2) + B^(n1)}/F]. Since F^2, (C  B)/F and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F have a highest common factor = 1 and if Fermat’s Last Theorem is false then F^2 (C  B)/F[{C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)}/F] = Aⁿ means that F^2 = G^n, (C  B)/F = H^n and [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n where G, H and I are positive integers. Consider [C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1)]/F = I^n. C^(n1) + C^(n2) B + ….. + C B^(n2) + B^(n1) = FI^n. C^(n2) B + C^(n3) B^2 + ….. C^2 B^(n3) + C B^(n2) = FI^n  B^(n1)  C^(n1) CB[C^(n3) + C^(n4) B + ….. C B ^(n4) + B^(n3)] = FI^n  B^(n1)  C^(n1). This implies that FI^n  B^(n1)  C^(n1) must be a multiple of C and B simultaneously. This is a contradiction because FI^n  B^(n1)  C^(n1) cannot simultaneously be a multiple of B and C. This is because if FI^n  B^(n1) is a multiple of C it cannot be a multiple of A or B and therefore FI^n  B^(n1)  C^(n1) which will be a multiple of C if FI^n  B^(n1) is a multiple of C cannot also be a multiple of A or B since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. Similarly if FI^n  C^(n1) is a multiple of B it cannot be a multiple of A or C and therefore F I^n  B^(n1)  C^(n1) which will be a multiple of B if FI^n  C^(n1) is a multiple of B cannot also be a multiple of A or C since FI^n, B and C have a highest common factor f = 1 and FI^n is a factor of Aⁿ. The only way for FI^n  B^(n1)  C^(n1) to simultaneously be a multiple of B and C is if B, C and therefore A have a highest common factor > 1. This proves the validity of Beal’s conjecture. 

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