Proving the Diophantine Equation x^3 + y^3 = 3z^3 Has No Integer Solutions

[gonzo]

I think I posted this in the wrong forum before. Let's try again.

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?

 

[robert Ihnot]

Well, one thing is sure 3 divides (X+Y), and it follows that 3 divides X^2-XY+Y^2, by modulo arithmetic, but only once. Proof:

Y==-X Mod 9 implies X^2-X(-X)+(-X)^2==3X^2 Mod 9. But 3 divides X implies 3 divides Y and Z. So we can remove such a factor.

So that we have X+Y =(3a)^3, X^2-XY+Y^2=3b^3. Z=3ab.

Well, maybe that helps...

 

[gonzo]

The only problem is if z has a factor of 3. We can then write:

$x^3 + y^3 = 3^4z'^3$

If we then set:

$p=(x+y)$
$q=(x^2-xy+y^2)$

Then we have 4 factors of 3 to divide up between p and q. It is easy to show that all options are impossible except one. 27 divides p and 3 divides q.

All cases except this one are easy. This is what I am looking for some insight on (or a completely different approach that bypasses the need to divide it up like this).

 

[malaygoel]

The only problem is if z has a factor of 3. We can then write:

$x^3 + y^3 = 3^4z'^3$

If we then set:

$p=(x+y)$
$q=(x^2-xy+y^2)$

Then we have 4 factors of 3 to divide up between p and q. It is easy to show that all options are impossible except one. 27 divides p and 3 divides q.

All cases except this one are easy. This is what I am looking for some insight on (or a completely different approach that bypasses the need to divide it up like this).

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

 

[gonzo]

You can't set composite factors equal like that.

 

[MalayInd]

you can substitute
x+y=3a
x-y=b
it will simplify your calculations
try it please

 

[murshid_islam]

check out the chapter (if you have not yet done it) on "some diophantine equations" in Hardy's "An Introduction to the Theory of Numbers". there is a proof of exactly what you need, gonzo.

 

[gonzo]

Thanks, but I don't have that book and I doubt it's in my library. I found a solution anyway using infinite descent.

 

[robert Ihnot]

Ribenboim points out that Lagrange was the first to show no solutions to $$X^3+Y^3=3Z^3$$, and that much work has been done on the equation: $$X^3+Y^3=aZ^3$$, though he did not give details.

IA Barnett, Elements of Number Theory points out that no solution for $$X^P+Y^P=PZ^P$$ exists, P>2, unless P divides Z. (This is relatively easy to show) HOWEVER, nothing is said about the case where P DOES DIVIDE Z, and I am left wondering if this matter is largely unknown.