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
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