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proof of fermat-catalan conjecture through the ... - Nardelli - Xoom.it

proof of fermat-catalan conjecture through the ... - Nardelli - Xoom.it

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Versione 1.022/3/2013Pagina 35 di 495. PROOF OF TIJDEMAN'S CONJECTURETijdeman's <strong>conjecture</strong> states that <strong>the</strong>re are at most a fin<strong>it</strong>e number <strong>of</strong> consecutive powers.Stated ano<strong>the</strong>r way, <strong>the</strong> set <strong>of</strong> solutions in integers b, c, n, k <strong>of</strong> <strong>the</strong> exponential diophantine equationfor exponents n and k ≥ 2, is fin<strong>it</strong>e.A + b n = c kIf A =1 <strong>the</strong> <strong>the</strong>orem was proven by Mihailescu and <strong>it</strong>'s <strong>the</strong> Catalan's <strong>conjecture</strong>, as we have seen beforeand is also known as Tijdeman's <strong>the</strong>orem.But w<strong>it</strong>h A ≥ 2, n and k ≥ 2 we have an unsolved problem, called <strong>the</strong> generalized Tijdeman problem. Itis <strong>conjecture</strong>d that this set also will be fin<strong>it</strong>e. This would follow from a yet stronger <strong>conjecture</strong> <strong>of</strong> Pillaistating that <strong>the</strong> equationA + Bb n = Cc konly has a fin<strong>it</strong>e number <strong>of</strong> solutions, that we have shown above.In factLet’s apply <strong>the</strong> new abc <strong>conjecture</strong>rad (A*b n c k ) ≤ Abc < Ac n kcc k < (A) 62 2π πc 6⎛ k ⎞⎜ + 1⎟ ⎝ n ⎠Now to solve this inequal<strong>it</strong>y we assume that2πA 6= cπ2e6

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