A diophantine equation - Page 2

DMT · #16 $Old$ June 1st, 2006, 09:52 AM

Quote:

Originally Posted by ThePerfectHacker

Maybe this would help.
$x,y$ must be odd and $z$ is even.

That's not true. You only know exactly one of the 3 is even, but it could be either one. Unless you found a good proof to the contrary?

Anyway, it doesn't matter, like I said, I have a solution that doesn't depend on parity.

ThePerfectHacker · #17 $Old$ June 1st, 2006, 10:05 AM

Quote:

Originally Posted by DMT

That's not true. You only know exactly one of the 3 is even, but it could be either one. Unless you found a good proof to the contrary?

I found it by contradiction, that is only possible case.
By considers the integers, $4q+r$

Quote:

Originally Posted by DMT

Anyway, it doesn't matter, like I said, I have a solution that doesn't depend on parity.

I was hoping you find a soltion via infinite descent and post it here. But alas, it is difficult.
----
Did you use Kummer's method/theorem?

DMT · #18 $Old$ June 1st, 2006, 10:21 AM

I posted the start of the infiinite descent solution above. It should be fairly easy from there. I can give more details if needed, but it's not that hard to work out starting from where I showed.

malaygoel · #19 $Old$ June 21st, 2006, 09:52 AM

Quote:

Originally Posted by DMT

For those who care, I do have a solution now, working of course in $Q(\zeta)$ where zeta is the third root of unity. We can then easily show that:

$x+y=27a^3$
$x+y\zeta = j(1-\zeta)(u+v\zeta)^3$

where j is a unit. We can ignore negative units easy enough, which leave three cases to try:

$j=\zeta$
$j=\zeta^2$
$j=1$

The first two lead to a direct contradiction, and the third leads to an infinite descent step.

Could someone please explain to me this solution.

Keep Smiling
Malay

ThePerfectHacker · #20 $Old$ June 21st, 2006, 09:56 AM

Quote:

Originally Posted by malaygoel

Could someone please explain to me this solution.

Keep Smiling
Malay

No offense malay, but you are no experienced enough to follow these concepts. Not because you are not cabaple but rather because you have never studied them.

You need to be familiar with, Algbraic Number Theory. Which itself you need to be familiar with Abstract Algebra and Field Theory.

malaygoel · #21 $Old$ June 21st, 2006, 10:00 AM

Quote:

Originally Posted by DMT

I need to prove that the equation
$x^3 + y^3 = 3z^3$

has no integer solutions. I can do it easily for all cases except where z has a factor of 3, in which case I don't know what to do.

I am assuming the 3 in front of the z term is supposed to make this easier somehow than the same equation without it, but I'm failing to see the simplification that this allows.

Anyone know?

The equation can be rewritten as
$z^3 = (x^3 - z^3) + (y^3 - z^3)$
Here, $z^3$ is expressed as the sum of two integers $a= x^3 - z^3$ and $b= y^3 - z^3$ which I claim is impossible.
I could restate the above statement in the following way:
I claim that it is impossible to break a perfect cube $z^3(= a+b)$ in two integers a and b such that $z^3 + a(= x^3)$ as well as $z^3 +b(=y^3)$ , both are perfect cubes.
Hence there are no solutions to the given equation.
The claim can be easily proved.
Keep Smiling
Malay

malaygoel · #22 $Old$ June 21st, 2006, 10:08 AM

Quote:

Originally Posted by malaygoel

I claim that it is impossible to break a perfect cube $z^3(= a+b)$ in two integers a and b such that $z^3 + a(= x^3)$ as well as $z^3 +b(=y^3)$ , both are perfect cubes.
Hence there are no solutions to the given equation.
The claim can be easily proved.

Let $z^3 = a + b$
and $z^3 + a= (z+k)^3$
hence, $a= k^3 + 3zk(z+k)$
$b=z^3 - a=z^3 - k^3 - 3kz(z+k)$
Now, $z^3 +b = 2z^3 - k^3 - 3zk(z+k)$
which cannot be a perfect cube for any value of k.
Hence the clain is proved and the question is solved.

Keep Smiling

Malay

DMT · #23 $Old$ June 22nd, 2006, 01:39 AM

Quote:

Originally Posted by malaygoel

Could someone please explain to me this solution.

Keep Smiling
Malay

It's a real pain to type out in a forum. But it's pretty simple with the information I posted if you understand basic algebraic number theory. If you tell me where you are getting stuck I could help a bit if you give it a try.

DMT · #24 $Old$ June 22nd, 2006, 01:42 AM

Quote:

Originally Posted by malaygoel

Let $z^3 = a + b$
and $z^3 + a= (z+k)^3$
hence, $a= k^3 + 3zk(z+k)$
$b=z^3 - a=z^3 - k^3 - 3kz(z+k)$
Now, $z^3 +b = 2z^3 - k^3 - 3zk(z+k)$
which cannot be a perfect cube for any value of k.
Hence the clain is proved and the question is solved.

Keep Smiling

Malay

You do realize this doesn't make sense, right? And even if you did manage to show that FLT for p=3 implies $2z^3 - k^3 - 3zk(z+k)$ is a perfect cube for integers z and k, you haven't proven that this is impossible, just stated it.

malaygoel · #25 $Old$ June 22nd, 2006, 04:59 AM

Quote:

Originally Posted by DMT

You do realize this doesn't make sense, right? And even if you did manage to show that FLT for p=3 implies $2z^3 - k^3 - 3zk(z+k)$ is a perfect cube for integers z and k, you haven't proven that this is impossible, just stated it.

Could you show that $2z^3 - k^3 - 3zk(z+k)$ is a perfect cube for any positive integer z and integer k?
Let $S=2z^3 - k^3 - 3zk(z+k)$
Let z be a given natural number.
I will prove that there is no integer k(k is an integer for which $z^3 + a$ is a perfect cube) such that S is a perfect cube.
$S = z^3 + (z-k)^3 - 6zk^2$
S is a perfect cube if and only if $-6zk^2 = 3z(z-k)(2z-k)$
that is, S is a perfect cube iff $3k^2 -3kz + 2z^2=0$
which has no solutions.
Hence , S can never be a perfect cube.
Keep Smiling
Malay

DMT · #26 $Old$ June 22nd, 2006, 08:54 AM

Quote:

Originally Posted by malaygoel

Could you show that $2z^3 - k^3 - 3zk(z+k)$ is a perfect cube for any positive integer z and integer k?

I assume you understand that your comment here is utterly meaningless? I'm not sure what point you are trying to make.

Quote:

Originally Posted by malaygoel

$S = z^3 + (z-k)^3 - 6zk^2$
S is a perfect cube if and only if $-6zk^2 = 3z(z-k)(2z-k)$

How exactly did you make this step?

malaygoel · #27 $Old$ June 22nd, 2006, 07:51 PM

Quote:

Originally Posted by DMT

How exactly did you make this step?

I have just completed the cube.
You would have sometimes completed the square.
$(a+b)^2 = a^2 + b^2 +2ab$
If you an expression, $ax^2 + bx + c$ and you want to find in what conditions it is a perfect square, we would do some adjustments and express it as $a(x + \frac{b}{2a})^2 + \frac{4ac-b^2}{4a}.$
For it to be a perfect square, $b^2 = 4ac$ and $a$ should be a perfect square.
I have done exactly the same thing, I have compared S with $(a+b)^3$ and equated the term $-6zk^2$ with $3ab(a+b)$ .
You can see it in this way also.
$S=(2z-k)^3 - 3z(3k^2 - 3kz + 2z^2)$
For S to be a perfect cube
$3k^2 - 3kz + 2z^2$ has to be zero for some value of k for a given z which is impossible.

Keep Smiling
Malay

DMT · #28 $Old$ June 23rd, 2006, 01:17 AM

Quote:

Originally Posted by malaygoel

You can see it in this way also.
$S=(2z-k)^3 - 3z(3k^2 - 3kz + 2z^2)$
For S to be a perfect cube
$3k^2 - 3kz + 2z^2$ has to be zero for some value of k for a given z which is impossible.

Keep Smiling
Malay

That's not necessarily true. You have stated this without showing it.

Why does that term have to be 0? For example, you have:

-8=64-72

-8 is a cube, 64 is a cube, and 72 is divisible by 3, so at first quick glance you can't rule it out, and even if you do rule it out, you haven't shown that nothing works.

You keep doing this Malay. You seem to misunderstand a fundamental idea about proofs and assumptions. I would suggest you spend some time improving that understanding before continuing with this line of attack or you will likely continue to make the same type of error over and over again.

The p=3 case of FLT does not have a simple solution (though it is simpler than most of the higher order cases). I don't know of any that don't use infinite descent.

malaygoel · #29 $Old$ June 23rd, 2006, 05:19 AM

Quote:

Originally Posted by malaygoel

I have just completed the cube.
You would have sometimes completed the square.
$(a+b)^2 = a^2 + b^2 +2ab$
If you an expression, $ax^2 + bx + c$ and you want to find in what conditions it is a perfect square, we would do some adjustments and express it as $a(x + \frac{b}{2a})^2 + \frac{4ac-b^2}{4a}.$
For it to be a perfect square, $b^2 = 4ac$ and $a$ should be a perfect square.
I have done exactly the same thing, I have compared S with $(a+b)^3$ and equated the term $-6zk^2$ with $3ab(a+b)(a=z,b=z-k)$ .

Is there anything wrong in this work?

KeepSmiling
Malay