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Versione 1.022/3/2013Pagina 32 di 49soc k < cπ2e62πc 6⎛ k ⎞⎜ + 1⎟ ⎝ n ⎠2πc k < c 6⎛ k ⎞⎜ + 1+e⎟ ⎝ n ⎠This implies that we have no solutions ifk ≥2π6⎛ k ⎞⎜ + 1 + e⎟⎝ n ⎠and hence dividing by k1 ≥2π6⎛ 1 1 e ⎞⎜ + + ⎟⎝ n k k ⎠It depends on values of n, k and e⎛ 1 1 e⎜ + +⎝ n k k⎞⎟⎠6≤ 2π= 0,6079…and hence we have solutions if and only if61⎝ n1kek2π = 0,6079… < ⎛ ⎞⎜ + + ⎟ ⎠< 1withe ≥ 0
Versione 1.022/3/2013Pagina 33 di 49if e=0 we have:61 1+⎝ n k2π = 0,6079… < ⎛ ⎞⎜ ⎟ ⎠< 12 2 2π π πB 6 C−16(A 6) = cπ2e6and we have the case of Catalan conjecture with A = 1, B = 1 and C =1, in fact:2 2 2π π π1 6 1−16(1 6) = c 0 = 1instead if e=1 we have:61 2+⎝ n k2π = 0,6079… < ⎛ ⎞⎜ ⎟ ⎠< 1a special case with2 2 2π π πB 6 C−16(A 62π) = c 6Of course it must always apply the inequality1 1 1+ + < 1m n kso e is limited by n and k
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