Ola' Albert e pessoal da lista, complementando o assunto, segue um link bonitinho...
http://www.eleves.ens.fr/home/baglio/maths/26number.pdf []'s Rogerio Ponce 2009/4/10 Albert Bouskela <bousk...@ymail.com>: > Olá! > > > > Esses alunos... > > > > Sua dileta aluna andou lendo sobre uma das mais engenhosas demonstrações de > Fermat. > > > > É verdade: 26 é o único inteiro compreendido entre um quadrado (25 = 5^2) e > um cubo (27 = 3^3). > > > > Formalmente, Fermat (que não era muito chegado a uma formalidade) demonstrou > que: > > > > |m^3 – n^2| > 2 para “m” e “n” inteiros, tais que m>3 e n>5 . > > > > Infelizmente, não achei exatamente a prova de Fermat na Internet, mas, > certamente, quem procurá-la com mais afinco, deverá encontrá-la. A prova que > achei não está completa – veja-a abaixo: > > > > http://abstractnonsense.wordpress.com/2006/08/28/algebraic-number-theory/ > > > > Algebraic Number Theory > > After explaining one elementary technique in number theory, I should write > about what motivates some of the basic ideas of algebraic number theory by > means of a somewhat more complicated proof, namely that 26 is the only > integer sandwiched between a square and a cube. > > > > In order to find other numbers similarly sandwiched, we need to solve each > of the equations x^2 + 2 = y^3 and x^2 - 2 = y^3. Apart from a few > degenerate solutions in which x or y is zero, we only know one integer > solution: x = +/-5, y = 3, which corresponds to 25 and 27. > > > > This time, we can’t take quadratic residues, because of that pesky third > power. All we can do is tell that x and y are odd; if one is even and one is > odd, then the equations say that an odd number and an even number are equal, > whereas if they’re both even, then we have a problem since y^3 is divisible > by 8, whereas x^2 +/- 2 isn’t even divisible by 4. > > > > It would be great if we could factor the left-hand side… which is a problem, > since neither 2 nor -2 is a perfect square. But let’s forget about that > hurdle for the moment and try factoring anyway. > > > > We have x^2 + 2 = (x + SQRT(-2))(x - SQRT(-2)). So instead of working just > with regular integers - which I’ll call rational integers because they’re > all rational numbers - we can work with regular integers, plus the square > root of -2. In particular, we work with the set {a + b*SQRT(-2): a and b are > integers}, consisting of numbers like 5, 3 + SQRT(-2), -3 - 4SQRT(-2), etc. > Since it’s possible to add, subtract, and multiply numbers like this > normally, this set forms a ring. > > > > Now, let’s look at the two factors, (x + SQRT(-2)) and (x - SQRT(-2)), a > little more closely. In particular, let’s look at any common divisors they > have, except the trivial ones 1 and -1. Any common divisor will have to > divide their difference, 2SQRT(-2) = -SQRT(-2)^3. So this common divisor is > SQRT(-2), 2, or 2SQRT(-2), which is divisible by SQRT(-2). > > > > That means that x + SQRT(-2) is divisible by SQRT(-2), or, if you will, that > x is divisible by SQRT(-2). But x/SQRT(-2) = (x/2)SQRT(-2), and we’ve > already proven that x is odd, so there’s a contradiction, and the two > factors have no common divisors. > > > > If they have no common divisors, then they’re both cubes. This is fairly > common sensical: any prime factor that divides the first factor has to > divide y^3. So its cube must divide y^3, too, which means it divides the > first factor, or else the first and second factor are both divisible by that > prime. > > > > So there’s a number, call it a + bSQRT(-2), such that (a + bSQRT(-2))^3 = x > + SQRT(-2). Expanding the left-hand side, we get that a^3 + 3a^2*bSQRT(-2) - > 6ab^2 - 2b^3*SQRT(-2) = x + SQRT(-2). Both the rational-integer and the > SQRT(-2) parts must be equal, so we have 3a^2*b - 2b^3 = 1, where a and b > are rational integers. > > > > Now we have enough to apply simpler tricks. The left-hand side is divisible > by b, so b has to be +/-1. If it’s -1, then we get -3a^2 + 2 = 1, or 3a^2 = > 1, which is absurd since a is a rational integer. If b = 1, then we have > 3a^2 - 2 = 1, or 3a^2 = 3, which means a = +/-1. > > > > If a = 1, then (a + SQRT(-2))^3 = -5 + SQRT(-2), so x = 5. Similarly, if a = > -1, then x = -5. Then y = 3 and we get 26. > > > > We can do exactly the same thing with the other equation, only this time we > work with SQRT(2). All the steps work exactly the same, only we end up with > 3a^2*b + 2b^3 = 1. In that case, b = 1 gives 3a^2 = -1, a contradiction, and > b = -1 gives 3a^2 = -3, another contradiction. > > > > So 26 is really the only number sandwiched between a square and a cube… > supposedly. I say “supposedly” because I lied to you a bit - actually, > there’s one or two very important things left to check that I didn’t check > here. In this case they work, but they don’t have to, and I need to show > that they work. But that’s for next time. > > > > Usando ferramentas mais “pesadas” do que as que Fermat conhecia, a prova > fica mais enxuta: > > > > http://www.mathhelpforum.com/math-help/number-theory/33404-proof-26-only-number-between-cubed-squared-number.html > > > > Sds., > > AB > > ________________________________ > Veja quais são os assuntos do momento no Yahoo! + Buscados: Top 10 - > Celebridades - Música - Esportes ========================================================================= Instruções para entrar na lista, sair da lista e usar a lista em http://www.mat.puc-rio.br/~obmlistas/obm-l.html =========================================================================
