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From: John on 28 May 2010 10:30
Prove that there is no natural x,y,z such that x^2 z + y^2 x = z^2 y.
Can you give me a hint ?
From: David Rusin on 28 May 2010 14:00
In article <c0084902-9751-4e9f-ac7c-f81d007164fe(a)m21g2000vbr.googlegroups.com>,
John <to1mmy2(a)yahoo.com> wrote:
>Prove that there is no natural x,y,z such that x^2 z + y^2 x = z^2 y.
>Can you give me a hint ?

A homogeneous cubic equation in three variables (usually) describes an
elliptic curve; there are general methods which tend to work well (not
always) to discover all their solutions. In this case your cubic
equation reduces to an elliptic curve very closely related to
(it's "isogenous to") the elliptic curve that results from the case
n=3 of Fermat's Last Theorem (X^3+Y^3=Z^3). That means that "elementary"
proofs will exist, but I wouldn't expect them to be especially obvious.

Or, to put it another way, I could do a few more or less trivial
manipulations and show that from a solution (x,y,z) to your equation
I would be able to construct a solutions (X,Y,Z) to the Fermat
equation, and then I could just end by saying "... and it is well known
that no such solutions exist".

(Conversely if someone finds a very short solution to your question,
we could turn it around and get a short solution to the case n=3 of FLT!)

dave

From: David Bernier on 28 May 2010 18:02
David Rusin wrote:
> In article<c0084902-9751-4e9f-ac7c-f81d007164fe(a)m21g2000vbr.googlegroups.com>,
> John<to1mmy2(a)yahoo.com> wrote:
>> Prove that there is no natural x,y,z such that x^2 z + y^2 x = z^2 y.
>> Can you give me a hint ?
>
> A homogeneous cubic equation in three variables (usually) describes an
> elliptic curve; there are general methods which tend to work well (not
> always) to discover all their solutions. In this case your cubic
> equation reduces to an elliptic curve very closely related to
> (it's "isogenous to") the elliptic curve that results from the case
> n=3 of Fermat's Last Theorem (X^3+Y^3=Z^3). That means that "elementary"
> proofs will exist, but I wouldn't expect them to be especially obvious.
>
> Or, to put it another way, I could do a few more or less trivial
> manipulations and show that from a solution (x,y,z) to your equation
> I would be able to construct a solutions (X,Y,Z) to the Fermat
> equation, and then I could just end by saying "... and it is well known
> that no such solutions exist".
>
> (Conversely if someone finds a very short solution to your question,
> we could turn it around and get a short solution to the case n=3 of FLT!)

If n is a non-negative integer, is

x^3 + y^3 + z^3 = n (in integers) closely related to an elliptic curve?

Actually, I see it won't be homogeneous if n >= 1.


David Bernier

From: hagman on 28 May 2010 19:42
On 28 Mai, 16:30, John <to1m...(a)yahoo.com> wrote:
> Prove that there is no natural x,y,z such that x^2 z + y^2 x = z^2 y.
> Can you give me a hint ?

Let p be a prime and consider p-adic valuation v, i.e
v(x) = n if p^n | x but not p^(n+1) | x
If v(x)>0 then v(z^2 y) = v(x^2 z + y^2 x) > 0, i.e. one of v(y), v(z)
>0.
Similarly, v(y) > 0 or v(z) > 0 imply that one of the other v's is >0
We may assume that (x,y,z) is a primitive solution, i.e. at least one
(and hence
exactly one) of v(x), v(y), v(z) is 0.

If v(x)=0 then v(z) = v( z^2 y - x^2 z ) = v(y^2 x) = 2 v(y).
Similar v(y)=0 implies v(x) = v( x^2 z + y^2 x) = v(z^2 y) = 2 v(z)
and v(z)=0 implies v(y) = 2 v(x).
This implies that there are pairwise coprime numbers a,b,c such that
x = b^2*c, y = a*c^2, z = a^2*b (first only up to sign, but signs can
be hiddn in
a,b,c).
Indeed this yields
b^4 c^2 a^2 b + a^2 c^4 b^2 c = a^4 b^2 a c^2,
which simplifies to
b^3 + c^3 = a^3

From: David Bernier on 1 Jun 2010 00:00
David Bernier wrote:
> David Rusin wrote:
>> In
>> article<c0084902-9751-4e9f-ac7c-f81d007164fe(a)m21g2000vbr.googlegroups.com>,
>>
>> John<to1mmy2(a)yahoo.com> wrote:
>>> Prove that there is no natural x,y,z such that x^2 z + y^2 x = z^2 y.
>>> Can you give me a hint ?
>>
>> A homogeneous cubic equation in three variables (usually) describes an
>> elliptic curve; there are general methods which tend to work well (not
>> always) to discover all their solutions. In this case your cubic
>> equation reduces to an elliptic curve very closely related to
>> (it's "isogenous to") the elliptic curve that results from the case
>> n=3 of Fermat's Last Theorem (X^3+Y^3=Z^3). That means that "elementary"
>> proofs will exist, but I wouldn't expect them to be especially obvious.
>>
>> Or, to put it another way, I could do a few more or less trivial
>> manipulations and show that from a solution (x,y,z) to your equation
>> I would be able to construct a solutions (X,Y,Z) to the Fermat
>> equation, and then I could just end by saying "... and it is well known
>> that no such solutions exist".
>>
>> (Conversely if someone finds a very short solution to your question,
>> we could turn it around and get a short solution to the case n=3 of FLT!)
>
> If n is a non-negative integer, is
>
> x^3 + y^3 + z^3 = n (in integers) closely related to an elliptic curve?
>
> Actually, I see it won't be homogeneous if n >= 1.

B. Poonen wrote an article for the Notices of the AMS, March 2008
(vol. 55, number 3).

From the Introduction:

Is x^3 + y^3 + z^3 = n solvable in integers, given some n?

n = 29 -> solvable, 3^3 + 1^3 + 1^3 = 29.
n = 30 -> solvable, not known until 1999.
n = 33 -> ? (unsolved problem at the time of writing).

Poonen, B.: "Undecidability in Number Theory".

David Bernier


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