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

Versione 1.022/3/2013Pagina 5 di 492414 1,645 = 366.946,54 > 50411 1 1+ + = 0,976…m n k3 5 + 11 4 = 122 22π{rad (3*11*2*61)} 6> 148844026 1,645 = 851.172,04 > 148841 1 1+ + = 0,95m n k33 8 + 1549034 2 = 15613 3{rad (3*11*2*61*12697*13*1201)} 62π> 3805914951397798.107.238.786 1,645 = 37921958271956418245,68 > 380.591.495.13971 1 1+ + =0,95833…m n k1414 3 +2213459 2 = 65 7{rad (2*7*101*479*4621*5*13)} 62π> 4902227890625203.439.016.690 1,645 = 4002896288091211907,85 > 4.902.227.890.6251 1 1+ + = 0,976….m n k

Versione 1.022/3/2013Pagina 6 di 499262 3 + 15312283 2 = 113 7{rad (2*11*421*7*53*149*277*113)} 62π> 23526054804481716.025.927.261.498 1,645 = 5271498777020594764161,13 > 235.260.548.044.8171 1 1+ + =0,976….m n k17 7 +76271 3 = 21063928 2{rad (17*13*5867*2*79*33329)} 62π> 443689062789184(that is divisible for 16 and 64)6.827.909.123.074 1,645 = 1295398290051100057437,9 > 443.689.062.789.1841 1 1+ + = 0,976….m n kWe note that if 443689062789184 is divisible for 16 and 64, this value can be connected also with thefollowing Ramanujan function regarding the modes corresponding to the physical vibrations of thesuperstrings:8 =13⎡∞ cosπtxw'2−πxw'⎤⎢ ∫ e dx0⎥4⎢antilogcoshπx2⎥ ⋅πt− w'⎢4( ) ⎥⎣e φw'itw'⎦⎡ ⎛ 10 + 11 2 ⎞ ⎛ 10 + 7log⎢⎜ ⎟ + ⎜⎢⎣ ⎝ 4 ⎠ ⎝ 41422t w'2 ⎞⎤⎟⎥⎠⎥⎦.

Versione 1.022/3/2013Pagina 7 di 4943 8 + 96222 3 = 30042907 2{rad (43*2*3*7*29*79*109*275623)} 62π> 902576261010649124.303.909.686.222 1,645 = 153259525699804607160993,02 > 902.576.261.010.6491 1 1+ + = 0,95833….m n kThe first of these (1 m +2 3 =3 2 ) is the only solution where one of a, b or c is 1; this is the Catalanconjecture, proven in 2002 by Preda Mihailescu. Technically, this case leads infinitely many solutions(since we can pick any m for m>6), but for the purposes of the statement of the Fermat-Catalanconjecture we count all these solutions as one.PROOFa m + b n = c k1 1 1+ + < 1m n kAs the new abc conjecture is given:2π{rad (a m b n c k )} 6> c kLet’s applyrad (a m b n c k ) ≤ abc < c m kc n kc

Versione 1.022/3/2013Pagina 8 di 49Please not that we cannot write, as is as in the case of Fermat's Great Theorem the inequalityrad (a n b n c n ) < c 3becauseit doesn't applya < c and b < c.The values of a, b and c may be any as long as obviously > 12πc k < c 6⎛⎜⎝kmk ⎞+ + 1⎟ n ⎠This implies that we have no solutions ifk ≥2π6⎛⎜⎝kmk ⎞+ + 1⎟n ⎠and hence dividing by k1 ≥2π6⎛ 1 1 1 ⎞⎜ + + ⎟⎝ m n k ⎠It depends on values of m, n and k⎛ 1 1 1⎜ + +⎝ m n kand hence we have solutions if and only if6⎞⎟⎠6≤ 2π2π = 0,6079… < m n k= 0,6079…1 1 1+ + < 1

Versione 1.022/3/2013Pagina 9 di 49This is a stronger inequality of the initial statement1 1 1+ + < 1m n kwhich then is wrong, because for⎛ 1 1 1⎜ + +⎝ m n k⎞⎟⎠6≤ 2πwe cannot have integer solutions-and this is the lower limit.If1 1 1+ + > 1m n k= 0,6079…We have infinite solutions of one of (2, 2, k), (2, 3, 3), (2, 3, 4), (2, 3, 5).If1 1 1+ + = 1m n kWe have finitely many solutions of one of (2, 4, 4), (2, 3, 6), (3, 3, 3).In fact from the examples above we have seen that the MINIMUM value for⎛ 1 1 1 ⎞⎜ + + ⎟⎝ m n k ⎠was13 2 + 7 3 = 2 9

Versione 1.022/3/2013Pagina 10 di 492π{rad (13*7*2)} 6> 512182 1,645 = 5221,76 > 5121 1 1+ + = 0,9444….m n kAlso here, we note that 512 / 8 = 64 and that 8 is a Fibonacci’s number and the number ofthe modes corresponding to the physical vibrations of the superstrings by the followingRamanujan function:8 =13⎡∞ cosπtxw'2−πxw'⎤⎢ ∫ e dx0⎥4⎢antilogcoshπx2⎥ ⋅πt− w'⎢4( ) ⎥⎣e φw'itw'⎦⎡ ⎛ 10 + 11 2 ⎞ ⎛ 10 + 7log⎢⎜ ⎟ + ⎜⎢⎣ ⎝ 4 ⎠ ⎝ 41422t w'2 ⎞⎤⎟⎥⎠⎥⎦.

Versione 1.022/3/2013Pagina 13 di 49125)If we choose1 1 1 6+ + > 23 4 x π \x < 40,7then the only possible values are1 1 1 , …….5 6 40Please note that the value of 21 exceeds the upper limit1 1 1+ + < 1 and therefore isn’t good.m n k6)If we choose1 1 1 + + < 12 3 xx > 6then the only possible values are1 1 1 , …….7 8 NN →∞

Versione 1.022/3/2013Pagina 14 di 497)If we choose1 1 1 + + < 12 4 xx > 4then the only possible values are1 1 1 , …….5 6 NN →∞8)If we choose1 1 1 + + < 12 5 xx > 3,33then the only possible values are1 1 1 , …….4 6 NN →∞From the set of solutions above we have only solutions:in the conjecture of Fermat-Catalan when one of the exponents m or n or k is equal to 2.In our examples this happens almost always, except in the cases 1, 2, 3 and 5.In example 1)

Versione 1.022/3/2013Pagina 15 di 491 1 1 6+ + > 24 5 x π \x < 6,33then the only possible values are1 1 1 , and2 3 6So we can havea 4 + b 5 = c 3or any permutation of 3, 4 and 5, asa 3 + b 5 = c 4a 3 + b 4 = c 5The last equation cannot give solutions.In the same way we havea 5 + b 6 = c 4a 4 + b 6 = c 5a 4 + b 5 = c 6The last equation cannot give solutions.So we have the subseta 4 + b 5 = c 3a 3 + b 5 = c 4

Versione 1.022/3/2013Pagina 16 di 49a 5 + b 6 = c 4a 4 + b 6 = c 5The solutions for this quartet of equations is very complicated and can only be obtained with the aid ofcomputers.Also remember that we can choose any values for a, b and c, not the usual values "small" that we areaccustomed.So that we have solutions and, therefore, the difficulty lies in the exponential growth of these powers.

Versione 1.022/3/2013Pagina 17 di 492. PROOF OF THE CATALAN'S CONJECTURECatalan's conjecture (or Mihăilescu's theorem) is now a theorem that was conjectured by themathematician Eugene Charles Catalan in 1844 and proven in 2002 by Preda Mihăilescu.2 3 and 3 2 are two powers of natural numbers, whose values 8 and 9 respectively are consecutive. Thetheorem states that this is the only case of two consecutive powers. That is to say, that the onlysolution in the natural numbers offor c, k, b, n > 1c k − b n = 1is1 + b n = c k1 + 2 3 = 3 2so worth the inequality1 1 1+ + < 1m n kBut what value we have to choose for m?We use the trick of taking m = ∞ so that we have he following inequality1 ∞ + 2 3 = 3 21 1 + < 1n kLet’s apply the new abc conjecture and so let's show this conjecture:rad (1*b n c k ) ≤ bc < c n kc

Versione 1.022/3/2013Pagina 18 di 492πc k < c 6⎛ k ⎞⎜ + 1⎟ ⎝ n ⎠This implies that we have no solutions ifk ≥2π6⎛ k ⎞⎜ + 1⎟⎝ n ⎠and hence dividing by k1 ≥2π6⎛ 1 1 ⎞⎜ + ⎟⎝ n k ⎠It depends on values of n and k⎛ 1 1 ⎞⎜ + ⎟⎝ n k ⎠6≤ 2π= 0,6079…and hence we have solutions if and only if61 1⎝ n2π = 0,6079… < ⎛ ⎞⎜ + ⎟ k ⎠< 1This is a stronger inequality of the initial statement1 1 + < 1n kwhich then is wrong, because for

Versione 1.022/3/2013Pagina 19 di 49⎛ 1 1 ⎞⎜ + ⎟⎝ n k ⎠6≤ 2π= 0,6079…we cannot have integer solutions.We can prove this way:If n = 3 we have1 + b 3 = c khence we have solutions if and only if61 1⎝ n2π = 0,6079… < ⎛ ⎞⎜ + ⎟ k ⎠< 161 1+⎝ k2π = 0,6079… < ⎛ ⎞⎜ ⎟3 ⎠< 123πk < 218 − π= 3,6417…k > 23= 1,5and hence1, 5 < k < 3,6417…Having to take only integer valuesk = 2 ork = 3 ,this is impossible otherwise1 + b 3 = c 3Has no sense, the difference between two cubes is never equal to 1

Versione 1.022/3/2013Pagina 20 di 49therefore remains only this equation that is valid1 + b 3 = c 2with two unknowns b and cIs then the solution sought1 + 2 3 = 3 2But If we choose n = 4 we havehence we have solutions if and only if1 + b 4 = c k61 1⎝ n2π = 0,6079… < ⎛ ⎞⎜ + ⎟ k ⎠61 1+⎝ k2π = 0,6079… < ⎛ ⎞⎜ ⎟4 ⎠< 1< 124πk < 224 − π= 2,7938…k > 34= 1,3333…and hence1,3333…< k < 2,7938…Having to take only integer values

Versione 1.022/3/2013Pagina 21 di 4924πWe note that in the inequality k < 2 = 2,7938…, we have 24, number that is connected with the24 − πmodes corresponding to the physical vibrations of the bosonic strings by the following Ramanujanfunction:⎡∞ cosπtxw'2−πxw'⎤⎢ ∫ e dx0⎥4⎢antilog coshπx2⎥ ⋅πt− w'⎢4( ) ⎥=⎣e φw'itw'24⎦⎡ ⎛ 10 + 11 2 ⎞ ⎛ 10 + 7log⎢⎜ ⎟ + ⎜⎢⎣ ⎝ 4 ⎠ ⎝ 41422t w'2 ⎞⎤⎟⎥⎠⎥⎦.Furthermore, we note that 1,3333…< k < 2,7938… are very near to the following values: 1,33333333and 2,79256959, (the first is equal) values corresponding to frequencies concerning the musical systembased on the Phi (i.e. the aurea ratio 1,61803398…)k = 21 + b 4 = c 2with two unknowns b and cBut for what value of n we have that k = 2hence we have solutions if and only if1 + b n = c 261 1+⎝ 22π = 0,6079… < ⎛ ⎞⎜ ⎟n ⎠< 1

Versione 1.022/3/2013Pagina 22 di 4922πn < 212 −π= 9,2655….n > 2and hence2 < n < 9,2655….Having to take only integer values1 + b n = c 2n = 3, 4, 5, 6, 7, 8 or 9with two unknowns b and c.

Versione 1.022/3/2013Pagina 23 di 49Since the equation is symmetric1 + b n = c kis equivalent to solve the unknown n or kSo, for example, we LIMIT the field equations valid1 + b 2 3, 4, 5, 6, 7, 8, 9= c1 + b 3, 4, 5, 6, 7, 8, 9 = c 2Solving this set of equations of only 14 equations, we find the only possible solution is thefollowing1 + 2 3 = 3 2Thus, for example, it isn't possible to have an equation as1 + 10 3 = 2,68 7Apart from the fact that c is not an integer, what is important to note is the value of the exponent of2πthe inequality abc that exceeds the maximum value of = 1,64496(rad (10*2,68)) q = 2,68 7q = 2,098 > 1,6449that's no good and it shows that all the other infinite equations with any n and k aren't possible,except field equation above.Notes that the field equations have n exponent or k exponent equal to 2, otherwise we don't havesolutions.This occurs also in the general case of the conjecture of Fermat-Catalan where one of theexponents m or n or k is equal to 2.

Versione 1.022/3/2013Pagina 24 di 493. SUPER-GENERAL TRINOMIAL EQUATIONWe can also prove the case of super-general trinomial equationAa m + Bb q = Cc krad (A*B*C*a m b n c k ) ≤ ABCabc < ABC c m kc n kcCc k < (ABC) 6c k < (A 6Now to solve this inequality we assume that2 2π πc 62 2 2π π πB 6 C−16(A 6⎛⎜⎝) c 6km2π2 2 2π π πB 6 C−16k ⎞+ + 1⎟ n ⎠⎛⎜⎝km) = ck ⎞+ + 1⎟ n ⎠π2e6withsoe ≥ 0c k < cπ2e62πc 6⎛⎜⎝kmk ⎞+ + 1⎟ n ⎠2πc k < c 6⎛⎜⎝kmk ⎞+ + 1+e ⎟n ⎠

Versione 1.022/3/2013Pagina 25 di 49This implies that we have no solutions ifk ≥2π6⎛⎜⎝kmk ⎞+ + 1 + e⎟n ⎠and hence dividing by k1 ≥2π6⎛ 1 1 1 e ⎞⎜ + + + ⎟⎝ m n k k ⎠It depends on values of n, k and e⎛ 1 1 1 e⎜ + + +⎝ m n k k⎞⎟⎠6≤ 2π= 0,6079…and hence we have solutions if and only if61⎝ m1n1kek2π = 0,6079… < ⎛⎞⎜ + + + ⎟ ⎠< 1withe ≥ 061⎝ m1n1k2π = 0,6079… < ⎛ ⎞⎜ + + ⎟ ⎠< 12 2 2π π πB 6 C−16(A 6) = cπ2e6and we have the case of Fermat-Catalan conjecture with A = 1, B = 1 and C =1, in fact:

Versione 1.022/3/2013Pagina 26 di 492 2 2π π π1 6 1−16(1 6) = c 0 = 1instead if e=1 we have:61⎝ m1n2k2π = 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 kThis shows by the new abc conjecture that we have only finitely many solutions (a,b,c,m,n,k) where(m or n or k) = 2, one of the three, otherwise the inequality doesn't hold.--------------------------------------------------------------------------------------------------------------------------Famous example of RamanujanIn fact there are 2 solutionsa 3 + b 3 = 1729rad (a 3 * b 3 * 1729) > 172912 3 + 1 3 = 1729

Versione 1.022/3/2013Pagina 27 di 49rad (12 3 *1 3 * 1729) = 2*3*7*13*19 = 10374 > 1729or even better (for the quantity or “quality” of the radical)10 3 + 9 3 = 1729rad (10 3 * 9 3 * 1729) = 2*3*5*7*13*19 = 51870 > 1729

Versione 1.022/3/2013Pagina 28 di 493.1 BEAL'S CONJECTUREBeal's conjecture is a conjecture that statesIfa m + b q = c kwhere a, b, c, m, n, and k are positive integers with m, n, and k ≥ 3 then a, b, c have a common factor.ExamplesTo illustrate, the solution 3 3 + 6 3 = 3 5 has bases with a common factor of 3, and the solution 7 6 + 7 7 =98 3 has bases with a common factor of 7. Indeed the equation has infinitely many solutions, includingfor examplem m m m m m m m m+1[ a ( a + b )] + [ b( a + b )] = ( a + b )for any a , b,m > 3 . But no such solution of the equation is a counterexample to the conjecture, sincethe bases all have the factorm ma + b in common.The example 7 3 + 13 2 = 2 9 shows that the conjecture is false if one of the exponents is allowed to be 2.PropertiesA variation of the conjecture where m, n, and k (instead of a, b, c) must have a common prime factor isnot true. See, for example 27 4 + 162 3 = 9 7 (3, 4 and 7 have no common factor)By the case of super-general trinomial equationPROOF

Versione 1.022/3/2013Pagina 29 di 49Aa m + Bb q = Cc kWe have Beal’s conjecture replacingA = B = C =1And so we return to the Fermat–Catalan conjecture BUT with exponent m, n, and k ≥ 3 (2 is notallowed).This shows by the new abc conjecture that we have only finitely many solutions (a,b,c,m,n,k) where(m or n or k) = 2, one of the three, otherwise the inequality doesn't hold (remember that a, b, c mustbe coprime).In fact we have trivial solutions with a common factor only with m, n, and k ≥ 3.

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

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

Versione 1.022/3/2013Pagina 34 di 4961⎝ n1kek2π = 0,6079… < ⎛ ⎞⎜ + + ⎟ ⎠< 1However. all the cases are a stronger inequality of the initial statement1 1 1+ + < 1m n kThis shows by the new abc conjecture that we have only finitely many solutions (b,c,n,k) with (n,k) ≠(2,2), otherwise the inequality doesn't hold.

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

Versione 1.022/3/2013Pagina 36 di 49withe ≥ 0soc k < cπ2e62πc 6⎛ k ⎞⎜ + 1⎟ ⎝ n ⎠and then follow all the calculation as beforeand hence we have solutions if and only if61⎝ n1kek2π = 0,6079… < ⎛ ⎞⎜ + + ⎟ ⎠< 1withe ≥ 0if e=0 we have:1 1n⎛ ⎞6⎜ ⎟2π = 0,6079… < ⎝+ k ⎠< 12πA 6= cπ2e6and we have the case of Catalan conjecture with A = 1, in fact:2π(1 6 ) = c0= 1

Versione 1.022/3/2013Pagina 40 di 49Y2 2 2 1 41( x12, x23,x13)2 ∫ d w2 2π − − −=2x1wx2wx3w,x2ijij2= x − x . (5.11)This integral is known in closed form. Forx < it is equal to2, x 2x 212 23 13YThence, we have that:22 2 2 1 ⎡π⎛1+s − t − A ⎞ ⎛1−s + t − A ⎞( x x , x ) = − 2Li⎜ ⎟ − 2Li⎜ ⎟ +12, 23 13 222x13⎝ 2 ⎠A⎢⎣ 3⎝ 2 ⎠⎛1+s − t − A ⎞ ⎛1−s + t − A ⎞⎤− ln s lnt+ 2ln⎜⎟ln⎜⎟⎥⎝ 2 ⎠ ⎝ 2 ⎠⎦22x12x23s = , t = , A = ( 1−s − t) 2 − 4st. (5.12)22x xY13132 2 2 1 41( x , x , x ) d w=122313=2 ∫ 2 2 2π x − w x − w x − w21 ⎡π⎛1+s − t − A ⎞ ⎛1−s + t − A ⎞= ⎢ − 2Li2⎜⎟ − 2Li2⎜⎟ +2x13A ⎣ 3 ⎝ 2 ⎠ ⎝ 2 ⎠⎛1+s − t − A ⎞ ⎛1−s + t − A ⎞⎤− ln s lnt+ 2ln⎜⎟ln⎜⎟⎥. (5.12b)⎝ 2 ⎠ ⎝ 2 ⎠⎦12This expression is valid for s , t < 1 with the principle branch of the logarithms and dilogarithms. If x12is the largest, then the result is the same function divided by s and the replacement s → 1/s and3t → t / s . Likewise when x23is the largest. In our case, if we take φ > 2π / 3, then for z ≤ 1 the firsttwo arguments of Y in (5.10) are less than unity and in that regime Y evaluates to2iφ−iφiφiφ−iφ⎞( − ze ) − Li ( 1+ze ) − log( z) log( 1+ze ) + log( − e ) log( + ze ) ⎟ 2i ⎛ πY = −⎜ − Li221zsinφ⎝ 63. (5.13)⎠The integration then gives∫( π −φ)( π + φ)1φdzY =. (5.14)0 3sinφWith the prefactor we find the final expression (valid by analytical continuation for all0 ≤ φ < π ) is

Versione 1.022/3/2013Pagina 41 di 49( 2 ) 4 cosθ− cosφVint( φ , θ ) =( π −φ)( π + φ)φ. (5.15)3 sinφThe first integral in (5.10) can also be done analytically. Again one should take care in choosingbranch cuts for the logarithms, where the principle branch is for small φ . The result is( 2 )( φ θ )22( cosθ− cosφ) ⎡⎛ π ⎞ i ⎤( ) ( ) ( ) ⎥⎦2iφ2iφ= −⎢ − −⎜ +⎟ + 3Vlad , 4Li23e ζ 3 iφLi2eφ . (5.16)sin φ ⎣⎝ 6 ⎠ 3Thence, form (5.10), we have the following expression:V( 2 )( φ θ )lad( cosθ− cosφ)2⎛ + ⎞ ⎛ += − ∫ ∞ iφidz 1 ze z elog⎜⎟log⎜20−iφ−sin φ z ⎝1+ze ⎠ ⎝ z + e,iφφ⎞⎟ =⎠22( cosθ− cosφ) ⎡⎛ π ⎞ i ⎤( ) ( ) ( ) ⎥⎦2iφ2iφ3= −4 ⎢Li3e − ζ 3 − iφ⎜ Li2e +⎟ + φ2sin φ ⎣⎝ 6 ⎠ 3. (5.16b)We have the following mathematical connection:− xxe −1 1−ek≥k∞∞∞−x−x−x( 1+k )∫ e dx = ∫ e dx = ∑∫ xe dx =−x−x∑0 0( cosθ− cosφ)0 0 ≥01( 1+k)2( ) ⎛ + ⎞ ⎛ +⇒ ( ) = − ∫ ∞ iφi2dz 1 ze z eVladφ θlog⎜⎟log⎜20−iφ−sin φ z ⎝1+ze ⎠ ⎝ z + e,iφ22π= ⇒622( cosθ− cosφ) ⎡⎛ π ⎞ i ⎤( ) ( ) ( ) ⎥⎦2iφ2iφ3= −4 ⎢Li3e − ζ 3 − iφ⎜ Li2e +⎟ + φ . (5.17)2sin φ ⎣⎝ 6 ⎠ 3φ⎞⎟ =⎠

Versione 1.022/3/2013Pagina 42 di 496. EXAMPLES OF PILLAI’S CONJECTUREWe see a table with x and y consecutive numbers with exponents 2 and 3, and then a similar table withexponents 3 and 4 for Pillaix^2 y^3 Algebraicdifferencey^3-x^2Square root ofthe difference3^2 = 9 2^3=8 1 1 14^2=16 3^3=27 11 3,31 15^2= 25 4^3= 64 39 6,24 96^2=36 5^3=125 89 9,43 97^2=49 6^3= 216 167 12,92 78^2=64 7^3=343 279 16,70 99^2=81 8^3=512 431 20,70 110^2=100 9^3=729 629 25,07 911^2=121 10^3=1000 879 29,64 912^2=144 11^3=1331 1187 34,45 713^2=169 12^3=1728 1559 39,48 914^2=196 13^3=2197 2001 44,73 115^2=225 14^3=2744 2519 50,18 9… … … … …Observationsfinal digits1, 7, 9 of thedifferencesAs we can see, the successive differences grow more and tend to infinity (their square roots growabout 4 or 5 at least up to x = 15), and therefore the only possible solution is the first,with one difference, as Catalan's conjecture. Furthermore, we note that the final digits of thedifferences are always 1, 7 and 9, are identified two successive groups of 9, 9, 7, 9, 1 after the pair 1, 7Initial.They are absent then the final digits 3 and 5.The fact that he chose:x 3 - (x + 1) 2 = kx 3 - x 2 - 2x -k - 1 = 0The solutions of this equation gives the values of x, being the divisors of the number (k – 1), by therule of Ruffini.

Versione 1.022/3/2013Pagina 43 di 49We can observe that the trinomialx 3 - x 2 - 2x - 1for any value of x results in an odd integer with the final digit that can only be 1, 7 or 9.This proves precisely for induction.Since Pillai’s conjecture applies to the equationA + Bb n = Cc kCc k - Bbn = Athe difference A is an integer odd or even that can also be repeated a finite number of times.In fact the sequence of the powers perfect is the following.1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343,361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, 1156, 1225, 1296,1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764Examples:2 5 – 3 3 = 512 2 – 2 7 = 16

Versione 1.022/3/2013Pagina 44 di 49Now we see the same table for the conjecture Pillai, limited to the exponents 3 and 4 in place of 2 and3.x^3 y^4 Algebraicdifferencey^4- x^3Square root ofthe difference3^3 = 27 2^4 = 16 11 3,31 14^3=64 3^4=81 17 4,12 75^3=125 4^4= 256 131 11,44 16^3= 216 5^4=625 409 20,22 97^3=343 6^4=1296 953 30,87 38^3=512 7^4=2401 1889 43,46 99^3= 729 8^4=4096 3367 58,02 710^3=1000 9^4=6561 5561 74,57 111^3=1331 10^4= 10000 8669 93,10 912^3=1728 11^4= 14641 12913 113,63 313^3=2197 12^4=20736 18539 136,15 914^3=2744 13^4=28561 25817 160,67 715^3=3375 14^4=38416 35041 187,19 1… … … … …Observationsfinal digits1, 3, 7 and 9 ofthe differencesNow, however, we note that the differences are still growing faster than in the table above, andtherefore also their square roots.Now the differences between the final digits there is also the number 3, but there is a little more orderin the sequences of the final digits: we identify two identical successive groups: 7 1 9 3 9:07 1 9 3 9after initial sequence 1; but also the third group begins with 7 and 1.Further calculations with successive powers and their differences (from Pillai's conjecture) show someinteresting regularities, which we quote:Consecutive powers (for conjecture Pillai)1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81,100,121,125,128,144, 169,196,216, 225,243, 256, 289, 324,343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000,1024, 1089, 1156,1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764

Versione 1.022/3/2013Pagina 45 di 49consecutive differences3 Fibonacci4 =2^21 only solution for the catalan’s conjecture (proved)7 prime9 =3^22 Fibonacci5 Fibonacci4 =2^213 Fibonacci15 =3*517 prime19 prime21 Fibonacci4 = 2^23 Fibonacci16 =4^225 =5^227 =3^320 =2*2*59 =3^218 =2*3*333 =3*1135 =5*719 prime18 =2*3*339 =3*1341 prime43 prime28 =2*2*717 prime47 prime49 = 7^251 prime53 prime55 Fibonacci57 =3*1959 = prime61 =prime39 =3*1324 =2*2*2*365 =5*13

Versione 1.022/3/2013Pagina 46 di 4967 prime69 = 3*2371 prime35 =5*738 = 2*1975 = 3*5*577 =7*779 prime81 =9^2= 3^447 prime36 =6^2We note that 4 occurs three times, twice on 3, 9 twice, 7 and 13 once, 6, 8, 10 and 12 zero times.The odd numbers in blue differ by 2 to groups of two, three or four consecutive numbers, with a groupof eight consecutive odd numbers. After this cycle falls to about half last number in blue (ex. 71/35 =2.02, 81/47 = 1.7) and after two or three sizes too small recurs again, obviously with consecutivenumbers gradually larger.They are present Fibonacci numbers (3, 5, 13, 21), square (4, 9, 16, 25, 36, 49, 81 and even a cube(27). We assume that this trend continues as the semiregular differences growth of consecutive powers,and could contribute to another definitive proof of Pillai's conjecture, according to which thesedifferences are repeated in a finite number of times.This means that the smaller numbers of this estimate will never be more later, and therefore theirpresence exists in a finite number of times, just as says the conjecture, so that proves true even if onlyby empirical means these numerical evidence, but not analytical.ConclusionsRemember that Pillai's conjecture states that for any given positive integer A there are only a finitenumber of pairs of perfect powers whose difference is A, but from the proof above that we have thatfor fixed positive integers A, B, C the equationCc k- Bb n = Ahas only finitely many solutions (b,c,n,k) with (n,k) ≠ (2,2).This is equivalent to saying that each positive integer A is valid and occurs only finitely many timesas a difference of perfect powers.

Versione 1.022/3/2013Pagina 47 di 496.1 PERFECT POWERS LINKED TO PILLAI’S CONJECTUREA sequence of perfect powers can be generated by iterating through the possible values for m and k.The first few ascending perfect powers in numerical order (showing duplicate powers) are thefollowing:2 2 = 4, 2 3 = 8, 3 2 = 9, 2 4 = 16, 4 2 = 16, 5 2 = 25, 3 3 = 27, 2 5 = 32, 6 2 = 36, 7 2 = 49, 2 6 = 64, 4 3 =64, 8 2 = 64,…The sum of the reciprocals of the perfect powers (including duplicates) is 1:∑∑∞ ∞m=2 k=21mk= 1.The first perfect powers without duplicates are:(sometimes 1), 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81,100,121,125,128,144, 169,196,216,225,243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900,961, 1000,1024, 1089, 1156, 1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764The sum of the reciprocals of the perfect powers p without duplicates is:1∑ ∑ ∞ =p k=p 2µ( k )( 1−ζ( k))≈ 0.874464365...where µ(k) is the Moebius function function and ζ(k) is the Riemann zeta function.Instead the sum of 1/(p−1) over the set of perfect powers p, excluding 1 and excluding duplicates, is 1:1 1 1 1 1 1 1 1∑ = + + + + + + + ... = 1.p p −13 7 8 15 24 26 31

Versione 1.022/3/2013Pagina 48 di 49This is sometimes known as the Goldbach-Euler theorem.But what happens if we add the differences of the reciprocals of the perfect powers?∑ ∞ 1= p k +− pk2 1k= 41 + 1 + 71 + 91 + 21 + 51 + 41 + 131 +….= ∞Since it must appear all natural numbers (with also duplicates) we have a sum a bit more of the seriesof the Harmonic series, that’s a divergent infinite series.This proves also that the gaps in the sequence of perfect powers tend to infinity.

Versione 1.022/3/2013Pagina 49 di 49References all on this web site http://nardelli.xoom.it/virgiliowizard/1) IL CONCETTO MATEMATICO DI “ABBONDANZA” E IL RELATIVOGRAFICO PER LA RH1Francesco Di Noto, Michele Nardelli (Gruppo B. Riemann)2) “NOVITA ‘ SULLA CONGETTURA DEBOLE DI GOLDBACH”Gruppo “B.Riemann”Francesco Di Noto,Michele Nardelli3)”Appunti sulla congettura abc”Gruppo “B. Riemann”**Gruppo amatoriale per la ricerca matematica sui numeri primi, sulle loro congetture e sulle loroconnessioni con le teorie di stringaFrancesco Di Noto, Michele Nardelli4) “I numeri primoriali p# alla base della dimostrazione definitiva della congettura di Goldbach(nuove evidenze numeriche)Francesco Di Noto, Michele Nardelli5) “ESTENSIONI DELLE CONGETTURE,FORTE E DEBOLE, DI GOLDBACH”(a k = primi , con N e k entrambi pari o dispari)Gruppo “B. Riemann”*Francesco Di Noto, Michele Nardelli*Gruppo amatoriale per la ricerca matematica sui numeri primi, sulle loro congetture e sulle loroconnessioni con le teorie di stringa.6) “IPOTESI SULLA VERITA’ DELLE CONGETTURE SUI NUMERI PRIMI CONGRAFICI COMET E CONTRO ESEMPI NULLI”(Legendre, Goldbach, Riemann…)Michele Nardelli ,Francesco Di Noto,