# A diophantine equation

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• May 28th 2006, 10:16 AM
DMT
A diophantine equation
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?
• May 29th 2006, 11:10 AM
ThePerfectHacker
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?

Let $S$ be the set of all positive integers such is if, $z\in S$ then it implies that there exists positive integers $x,y$ such as,
$x^3+y^3=3z^3$. Your aim is to show that this set is empty, i.e. the diophantine equation got no solution. Assume it is not then by the well-ordering principle there exists a minimal element in $S$ call it $z$. Then by definition of $S$ we have,
$x^3+y^3=3z^3$.

Note that the right hand side is divisible by 3. That means one of two things.
1)Both $x,y$ divisible by 3.
2)The integers $x,y$ leave remainders 1 and 2
---
If case 1) is true then we have,
$(3a)^3+(3b)^3=3z^3$
Thus,
$27a^3+27b^3=3z^3$
Thus,
$9a^3+9b^3=z^3$
Now the left hand side is divisible by 3, thus $3|z$ because 3 is prime. Thus,
$9a^3+9b^3=27c^3$
Thus,
$a^3+b^3=3c^3$ with $c which is impossible because of the minimal nature of $z$. (this reminds me of Fermat's infinite descent).
From this discussion we proved that condition 1) cannot be true.
---
If case 2) is true then we have, for some integers $k,j$ that,
$(3k+1)^3+(3j+2)^3=3z^3$ thus,
Open parantheses and organize,
$27(k^3+j^3)+27(k^2+2j^2)+9(k+4j)+9=3z^3$
Thus,
$9(k^3+j^3)+9(k^2+2j^2)+3(k+4j)+3=z^3$
Important note that the left hand side is divisible by 3, thus, $3|z^3$ thus, $3|z$ because 3 is prime. As a result we have that 3 divides $z$. This is your problem.
------------
I know I did not answer your question. I wrote it so that the next time I work on this problem I can continue from here.
• May 29th 2006, 01:43 PM
rgep
I don't have a complete solution either but here's a suggestion.

Try factoring in the ring Z[w] where w is a primitive cube root of unity. This ring is a PID. Assume there is no factor in common between x,y,z. We have (x+y)(x+wy)(x+w^2y) = 3z^3. If any two of x+y, x+wy, x+w^2y have an irreducible factor p in common, then p divides the difference, (w-1)y and so p divides y or w-1. If p divides y then p divides -x and hence z. So we assume p divides w-1: that is, p is sqrt(3). But if sqrt(3) divides two of these factors, it divides the third as the differences are all (unit multiples of) (w-1)y. Hence p^3 divides the LHS and so p divides z: that is, 3 divides z, so p^8 divides the RHS. Now p^6 divides one of the factors on the LHS, and p exactly divides the other two, as the difference (w-1)y is a multiple of sqrt(3) but not 3. If p^6 = 27 divides x+wy it divides x+w^2y as these terms are conjugate. So we have 3^3|x+y and 3^1 exactly divides (x+wy)(x+w^2y)=x^2+xy+y^2. This is pretty much the situation at the end of the previous posting.

Now apart from these factors of p, the terms on the LHS are coprime. Since their product is p^2 times a cube, each of them is a cube possibly times a power of p and a unit (power of w). If the integer x+y is a cube in Z[w] then it is a cube in Z. We get an equation of the form x+y = u^3, x+yw = (s+tw)^3 possibly with extra factors on the right of each.
• May 29th 2006, 02:25 PM
DMT
I got the first part of what you did ... basically reducing it to the case where 27 divides (x+y) and 3 divides (x^2-xy+y^2), but this is the case I couldn't rule out.

I had missed the coprime except for 3 part though ... since they had 3 in common I forgot they could be "otherwise coprime" that could be really helpful.
• May 29th 2006, 04:21 PM
ThePerfectHacker
Quote:

Originally Posted by rgep
I don't have a complete solution either but here's a suggestion.

Really? I always though you had the answers to every question they post here. I think this is the first time I seen you be confused.

I think diophantine equation are one of the most complicated problems in all of math. Since they are no standard way to solve them, furthermore some are even undecidable which is not the case here since it is not an exponential.
------
Quote:

Originally Posted by DMT
I got the first part of what you did ... basically reducing it to the case where 27 divides (x+y) and 3 divides (x^2-xy+y^2), but this is the case I couldn't rule out.

I had missed the coprime except for 3 part though ... since they had 3 in common I forgot they could be "otherwise coprime" that could be really helpful.

Not necessarily. That fact would be useful if you had,
$(x-y)(x^2-xy+y^2)=a^3$
Then because of relative primeness you can conclude,
$\left\{ \begin{array}{c}x-y=s^3\\x^2-xy+y^2=t^3$
Which is not the case here.
• May 30th 2006, 03:56 AM
DMT
Okay, I think I got it (and the coprime part IS essential by the way). The short version is you need to switch to ideals first and then work mod $<1-\omega>^3$ which gives you the key equation:

$x+y\omega = ra(1-\omega)\omega^n (mod <1-\omega>^3)$

where r is real and a is a rational integer. From there it is easy.
• May 30th 2006, 01:25 PM
ThePerfectHacker
I think I got it.
Read my first post to help understand how I am going to procede.
----
You have,
$x^3+y^3=3z^3$
and we have been able to settle that $x,y$ must have form $3k+1,3j+2$. Therefore, $3|(x+y)$.

I will now show that $\gcd (x,y)=1$. Assume otherwise, $\gcd(x,y)=d>1$, then,
$d^3(x')^3+d^3(y')^3=3z^3$
But, $d^3|3z^3$ and $d\not | 3$ thus, $d|z^3$ thus, $d|z$ thus,
$d^3(x')^3+d^3(y')^3=3d^3(z')^3$ thus,
$(x')^3+(y')^3=3(z')^3$, but $z' which cannot be becuase of the minimal nature of $z$ (read first post). Thus, $\gcd (x,y)=1$.

As you said correctly factor,
$(x+y)(x^2-xy+y^2)=3z^3$
but, $\gcd (x+y,x^2-xy+y^2)=1$ because $\gcd (x,y)=1$. Thus,
$\left( \frac{x+y}{3} \right) (x^2-xy+y^2)=z^3$
-=IMPORTANT=-
$\left( \frac{x+y}{3} \right)$ is an integer as explained in previous paragraph. Thus, (in post #5)
$\left\{ \begin{array} {c} \frac{x+y}{3}=a^3 \\ x^2-xy+y^2=b^3$
Thus,
$\left\{ \begin{array}{c} x+y=3a^3\\ x^2-xy+y^2=b^3$ Square equation 1 and from it subtract equation 2,
$3xy=9a^6-b^3$
Thus,
$b^3=9a^6-3xy$
Thus, $3|b^3$ thus, $3|b$. But then, $3|(x^2-xy+y^2)$ but $3|(x+y)$ thus, $\gcd (x+y,x^2-xy+y^2)\geq 3$
There are no solutions.
• May 30th 2006, 03:14 PM
DMT
First, my method turned out not to work ... I had made a sign error. I tried using Kummer's method despite the lack of restriction on 3 dividing Z, and thought it would work anyway, but it didn't when I fixed the sign error. Death by 3's.

You made an error midway PerfectHacker:

$
\gcd (x+y,x^2-xy+y^2)=1
$

This is not true, and this is the problem. These terms can share a factor of 3. Specifically, there is a problem is 3 is a factor of Z and 27 is a factor of x+y and the remaining 3 a factor of the quadratic term. This is the one case that can't be ruled out. This makes the rest of your proof incorrect since you are forgetting the other factors of 3.
• May 30th 2006, 06:46 PM
ThePerfectHacker
Quote:

Originally Posted by DMT
You made an error midway PerfectHacker

I knew it sounded to good to be true.
• May 30th 2006, 09:21 PM
malaygoel
A Simple Solution
Here I am providing a very simple solution(only for natural numbers). If you find any mistake, tell me.
The solution goes like this:
3z^3 = x^3 +y^3 =(x+y)(x^2 – xy + y^2)
Now two things to note here:
(1)z>3(you can check it yourself)
(2)The first factor on the right hand side of the equation(x+y) is less than the second(x^2 – xy + y^2)(I can prove it to you if you want)

Now I will factorize 3z^3 into two factors with smaller first:
(1) 1,3z^3
(2) 3,z^3
(3) 3z,z^2
(4) z,3z^2
(Are there any more factors?)
You can easily see first and second case are not applicable
(3) x+y=3z
(x^2 – xy + y^2)=z^2
Eliminating z we get
{(x+y)/3}^2 = (x^2 – xy + y^2)
(x^2 + 2xy + y^2) = 9(x^2 – xy + y^2)
8(x^2 – 2xy + y^2) = -5xy, .. … a contradiction
(4) x+y=z
(x^2 – xy + y^2)=3z^2 =3 (x+y)^2 =3(x^2 + 2xy + y^2)
2(x^2 – 2xy + y^2)= -9xy,…..a contradiction

All cases exhausted....
• May 31st 2006, 12:02 AM
DMT
Sorry Malaygoel, you can't set composite factors equal like that.
• May 31st 2006, 10:40 PM
malaygoel
Sorry for wasting your time but I have got something important now.
you can substitute
x+y=3a
x-y=b
• May 31st 2006, 11:18 PM
CaptainBlack
Quote:

Originally Posted by malaygoel
Here I am providing a very simple solution(only for natural numbers). If you find any mistake, tell me.
The solution goes like this:
3z^3 = x^3 +y^3 =(x+y)(x^2 – xy + y^2)
Now two things to note here:
(1)z>3(you can check it yourself)
(2)The first factor on the right hand side of the equation(x+y) is less than the second(x^2 – xy + y^2)(I can prove it to you if you want)

Now I will factorize 3z^3 into two factors with smaller first:
(1) 1,3z^3
(2) 3,z^3
(3) 3z,z^2
(4) z,3z^2
(Are there any more factors?)

If z is composite; yes.

RonL
• June 1st 2006, 05:52 AM
DMT
Quote:

Originally Posted by malaygoel
Sorry for wasting your time but I have got something important now.
you can substitute
x+y=3a
x-y=b

If you have some idea how to solve it, you should show it, but I doubt your suggestion will help.

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.
• June 1st 2006, 08:44 AM
ThePerfectHacker
Maybe this would help.
$x,y$ must be odd and $z$ is even.
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