Solve Fermat's Equation: Leopold Kronecker UFD

[Kummer]

Consider the Diophantine equation:
$$y^3 = x^2 + 2$$

Without using rational elliptic curves and unique factorization in $$\mathbb{Z}[\sqrt{-2}]$$ how many different ways can you show that this equation has only a single solution.


Historical question: Who was the mathematician who created the concept of UFD? I think it was Leopold Kroneckor, am I correct?

 

[Gib Z]

Perhaps look at the sequence of cubes and squares mod 4.

 

[ramsey2879]

Perhaps look at the sequence of cubes and squares mod 4.

No hope for that line of attact since (4n+3)^3 = 3 mod 4 which is 2 more than an odd square mod 4.
I narrowed it down to proving that the equation

$$y^3-27 = x^2 -25$$ has only two solutions in integers which is y = 3 and x = +/-5. Both sides can be factored but I don't know how to proceed after that. Perhaps mod 25

 

[Gib Z]

I don't understand, if you narrowed it down to proving the equation you posted has only 2 solutions in integers, and you found 2 solutions, aren't you done?

How did you narrow it down By the Way?

 

[ramsey2879]

I don't understand, if you narrowed it down to proving the equation you posted has only 2 solutions in integers, and you found 2 solutions, aren't you done?

How did you narrow it down By the Way?

I was already aware that 3^3 was 2 more than 5^2. But I am not sure that I have proven that there are only two solution sets to the equation, since each side is not necessarily equal to zero.

 

[ramsey2879]

I was already aware that 3^3 was 2 more than 5^2. But I am not sure that I have proven that there are only two solution sets to the equation, since each side is not necessarily equal to zero.

Further work
$$(3+m)^{3} = (5+n)^{2} + 2$$ expands to
$$27+27m + 9m^{2} + m^{3} = 25 + 10n + n^{2} + 2$$
$$27m + 9m^{2} + {m}^3 = 10n + n^{2}$$
$$m(27+9m + m^{2}) = n(10+n)$$
So we have to prove that m=0 and n=0 or -10 are the only solution in integers to the equation immediately above
lets consider it mod 4 with m = 0,1,2,3 mod 4 respectively to see if the left side is either 0 or 3 which are the only right side choices. If m = 0 mod 4 is the only fit then I think we made progress.

 

[Kummer]

The approaches that you are using are called Elementary solutions to diophantine equations. They are the best ones but often the most difficult methods. According to E.T. Bell, Euler has spend 7 years solving this equation with elementary techinques. This shows how difficult it is without advanced methods in number theory.

But I was not asking for an elementary approach, for I believe it is too long. I was asking for other approaches I have never seen before.

 

[robert Ihnot]

Kummer: Historical question: Who was the mathematician who created the concept of UFD? I think it was Leopold Kroneckor, am I correct?

I am surprised you would ask that! Wasn't it Kummer who thought he had solved Fermat's Last Theorem, until he discoverd the bit about unique factorization?


However, Wikipedia adds: "The extension of Kummer's ideas to the general case was accomplished independently by Kronecker and Dedekind during the next forty years." http://en.wikipedia.org/wiki/Ideal_number

And here is another tidbit from the same source:

"It is widely believed that Kummer was led to his "ideal complex numbers" by his interest in Fermat's Last Theorem; there is even a story often told that Kummer, like Lamé, believed he had proven Fermat's Last Theorem until Dirichlet told him his argument relied on unique factorization; but the story was first told by Kurt Hensel in 1910 and the evidence indicates it likely derives from a confusion by one of Hensel's sources. Harold Edwards says the belief that Kummer was mainly interested in Fermat's Last Theorem "is surely mistaken"

 

[Kummer]

I am surprised you would ask that! Wasn't it Kummer who thought he had solved Fermat's Last Theorem, until he discoverd the bit about unique factorization?

I assumed it was him but mostly his student, Leopold Kronecker that extended some ideas of Kummer ideals and Unique Factorization.

 

[rayb07]

I think I have a short proof:

http://rayb07.blogspot.com/2007/08/fermat-equation.html

 

[Kummer]

I think I have a short proof:

I do not understand it.

 

[rayb07]

Sorry, I don't have a proof, there's a simple error in it. My bad.

 

[ramsey2879]

Sorry, I don't have a proof, there's a simple error in it. My bad.

Not bad! It could be the new idea which Kummer asked for. Maybe, you just didn't follow it through properly.
You showed that if $$y^3$$ is an new candidate for Fermat's equation where $$y > 3$$ then there is an integer $$n> 0$$solving
$$n^{2} + 10n +27 -y^{3} = 0$$ [3]




You futher said [3] is a quadratic equation, and its coefficients can be labelled a, b, and c. Further, you noted that "In order for n to be an integer, b2 - 4ac must be a perfect square, k"



which reduces to $$k^{2} = 4(y^{3}-2)$$

Now $$y^{3}-2$$ is a square, call that square $$c^{2} | c>5$$

Then $$n = -5 + c | n>0$$ by the quadratic formula.

Pluging that back into your equation we get

$$25-10c +c^{2} -50+10c +27 -y^{3}$$ = 0
$$c^{2} = y^{3}-2$$

So your equation does check out but where to go after that I don't know

 

[robert Ihnot]

The matter of unique factorization in $$\sqrt-2$$ can be side stepped to a certain extent. But it involves some difficulity.

X==-2 Mod p is true for p =8k+1, and p=8k+3. The next matter is to show that those primes can be expressed in the form p=a^2+2b^2, and, hopefully, uniquely so. (Maybe that can be assumed to be written up somewhere.)

Then it is no trouble to show that numbers in the form of s^2+2t^2 are closed under multiplication.

Thus we arrive at the fact that Y^3 is in the form of (a^2+2b^2)^3. The next problem to discover is just how many different ways can this be resolved. For example even though 5 in the form a^2+b^2 can only be expresssed as 2^2 +1^2, 5^3 = 11^2+2^2, and 10^2+5^2.

Now if we find (a^2+2b^2)^3 = (a^3+2ab^2)^2 + 2[(a^2)b+2b^3]^2, this will NOT give the answer, but its fine if a=1, b=1 to find 3^3=3^2+2(3^2)--should that be of some use now and then.

BUT, if we can get (a^3-6ab^2)^2 + 2[3(a^2)b-2b^3]^2. THIS WILL WORK for a=b=1.
And...less I forget to say, it is an easy matter to see that the form: 2b^2[3a^2-2b^2] =2. can only be resolved for the case a, b = $$\pm1$$. (b doesn't matter and the answer for x is either $$\pm5$$.

 

[ramsey2879]

The matter of unique factorization in $$\sqrt-2$$ can be side stepped to a certain extent. But it involves some difficulity.

X==-2 Mod p is true for p =8k+1, and p=8k+3. The next matter is to show that those primes can be expressed in the form p=a^2+2b^2, and, hopefully, uniquely so. (Maybe that can be assumed to be written up somewhere.)

Then it is no trouble to show that numbers in the form of s^2+2t^2 are closed under multiplication.

Thus we arrive at the fact that Y^3 is in the form of (a^2+2b^2)^3. The next problem to discover is just how many different ways can this be resolved. For example even though 5 in the form a^2+b^2 can only be expresssed as 2^2 +1^2, 5^3 = 11^2+2^2, and 10^2+5^2.

Now if we find (a^2+2b^2)^3 = (a^3+2ab^2)^2 = 2[(a^2)b+2b^3]^2, this will NOT give the answer, but its fine if a=1, b=1 to find 3^3=3^2+2(3^2)--should that be of some use now and then.

BUT, if we can get (a^3-6ab^2)^2 + 2[3(a^2)b-2b^3]^2. THIS WILL WORK for a=b=1.
And...less I forget to say, it is an easy matter to see that the form: 2b^2[3a^2-2b^2] =2. can only be resolved for the case a, b = $$\pm1$$. (b doesn't matter and the answer for x is either $$\pm5$$.

I am just beginning to study fields and your insight is very much appreciated. I have something else I want to take a look at also:

Still another variable "w" to add to the mix.
Let A(0) = 0, A(1) = 1, A(n) = 2*A(n-1)-A(n-2) + w
If w = 1 then you have the triangular series
If w = 2 then you have the series of squares
In general $$A_n = w*(n^2 -n)/2 + n$$ is a direct formula for $$A_n$$

redefine $$D_{(b,n,m)$$ as:

$$(A_{(n-b)}-m)*(A_{(n+2b)}-m) - (A_{(n+b)}-m)*(A_{(n-2b)}-m)$$

Then
$$D_{(b,n,m)} = b(w*(2n-1)+2)*(A_{n} -wb^{2} -m)$$

Now if $$w=-b$$ and $$m = A_{n}+2$$ we get
$$\frac{D_{(b,n,m)}}{2b-b^2(2n-1)} = b^3-2$$
so the left hand side should be a square (25) only for b = 3 for all integer n.