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Versione 1.022/3/2013Pagina 30 di 493.1.1 ANOTHER PROOF OF FERMAT’S LAST THEOREM THROUGHBEAL’S CONJECTUREBeal's conjecture is a generalization of Fermat's last theorem, which corresponds to the casex x xx = y = z . If a + b = c with x ≥ 3 , then either the bases are coprime or share a common factor. Ifthey share a common factor, it can be divided out of each to yield an equation with smaller, coprimebases.Now since Beal's conjecture is true, Fermat’s last theorem can be proven by contradiction:Let there be positive integers n ≥ 3 and a, b, c such thatwhere c is the smallest possible.a n + b q = c nThen, by the Beal's conjecture with m = n = k = n it is shown there exists a base p dividing each of a,b and c. However, thenn⎛ a ⎞⎜ ⎟⎠⎝ p+n⎛ b ⎞⎜ ⎟⎠⎝ p=n⎛ c ⎞⎜ ⎟⎠⎝ pwhich contradicts C being the smallest possible, BECAUSE C DOESN’T HAVE TO BEDIVISIBLE BY p.This shows the positive integers n, a, b, c cannot exist, thus proving Fermat's last theorem.CVD
Versione 1.022/3/2013Pagina 31 di 494. PROOF OF PILLAI'S CONJECTUREPillai's conjecture concerns a general difference of perfect powers .Pillai conjectured that the gaps in the sequence of perfect powers tend to infinity.This is equivalent to saying that each positive integer occurs only finitely many times as a differenceof perfect powers: more generally, in 1931 Pillai conjectured that for fixed positive integers A, B, C theequationA + Bb n = Cc khas only finitely many solutions (b,c,n,k) with (n,k) ≠ (2,2).Let’s apply the new abc conjecturerad (A*B*C*b n c k ) ≤ ABCbc < ABCc n kcCc k < (ABC) 6c k < (A 6Now to solve this inequality we assume that(A 62 2π πc 62 2 2π π πB 6 C−162 2 2π π πB 6 C−16⎛ k ⎞⎜ + 1⎟ ⎝ n ⎠2π) c 6) = c⎛ k ⎞⎜ + 1⎟ ⎝ n ⎠π2e6withe ≥ 0
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