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Topic: 3x^3+4y^3+5z^3=0 (Read 5176 times) |
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Eigenray
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3x^3+4y^3+5z^3=0
« on: Jan 24th, 2008, 9:10am » |
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Show that the equation
3x3 + 4y3 + 5z3 = 0
has a non-trivial solution mod p for all primes p. In fact, show that it has a non-trivial solution mod any prime power. Conclude that it has a non-trivial solution mod n, for all integers n. If you know what this means: show that it has a non-trivial solution in the p-adics  p for all primes p.
Harder: Does it have a non-zero solution in  ? Hint: Code:> R<x>:=PolynomialRing(Integers());
> K:=NumberField(x^3-6);
> ClassNumber(K);
1
> IntegralBasis(K);
[
1,
K.1,
K.1^2
]
> G,m:=UnitGroup(K);
> G;
Abelian Group isomorphic to Z/2 + Z
Defined on 2 generators
Relations:
2*G.1 = 0
> m(G.1);
[-1, 0, 0]
> m(G.2);
[-1, 6, -3] |
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| « Last Edit: Jan 24th, 2008, 9:21am by Eigenray » |
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Michael Dagg
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Re: 3x^3+4y^3+5z^3=0
« Reply #1 on: Jan 24th, 2008, 3:10pm » |
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This is a classic example due I think to Serre. The point is
to show the failure of the local-to-global principle (a theorem which
says "there exist integer solutions iff there exist real and p-adic
solutions for all primes p"; that theorem applies to homogeneous
polynomials of degrees 1 and 2 but not, as this example shows, in
degree 3 or higher). You show the existence of p-adic solutions by
showing the existence of solutions mod p^n for all n; you can
pass from mod-p^n to mod-p^{n+1} solutions pretty easily for p not
equal to 3 (and p=3 is similar but a little harder because the
expansion of (x+ t p^n)^3 doesn't have a term linear in t ). So
it really comes down so solving the equation mod p, and I guess
that's because at least one of -3/4, -4/5, or -5/3 must be a
cube mod p . At least, that's the way I recall working this out once.
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| « Last Edit: Jan 24th, 2008, 3:18pm by Michael Dagg » |
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Eigenray
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Re: 3x^3+4y^3+5z^3=0
« Reply #2 on: Jan 24th, 2008, 4:41pm » |
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Actually it's due to Selmer, 1951, and was the first example of local-to-global failure in degree 3 (he considers the more general equation, but mentions this one specifically as the simplest).
However, the failure is somewhat finite in nature (analogous to how the class number provides a finite measure of the failure of unique factorization in the ring of integers of a number field): define the curve
C/Q : 3x3 + 4y3 + 5z3 = 0.
It turns out that there exist exactly 5 curves (up to isomorphism over Q) which are isomorphic to C over each Qp. That is, globally isomorphic here is only "finitely" stronger than locally isomorphic. For more, see, e.g., "On the passage from local to global in number theory" by Barry Mazur (the first third of which, at least, should be readable).
on Jan 24th, 2008, 3:10pm, Michael_Dagg wrote:| So it really comes down so solving the equation mod p, and I guess that's because at least one of -3/4, -4/5, or -5/3 must be a cube mod p . |
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There's a bit more to it than that. That argument doesn't work for p=13,31,61,...
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Michael Dagg
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Re: 3x^3+4y^3+5z^3=0
« Reply #3 on: Jan 25th, 2008, 9:08am » |
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Ah! I knew that -- I had my Se**** wrong.
> There's a bit more to it than that. That argument doesn't work
> for p=13,31,61,...
You're right, of course; I spoke too fast. When p = 2 mod 3, then every
element of Z/pZ is a cube and so the equation is easy to solve.
When p=1 mod 3, the multiplicative group (Z/pZ)^* is cyclic of order
(p-1), where multiplication by 3 is not a surjection, and I was
looking at the images of those three elements in the quotient group
(Z/(p-1)) / (3 Z/(p-1)) ~= Z/3Z
In this quotiient group, the three elements I proposed must sum to zero
but it is perfectly possible that the three of them are all congruent
(to something nonzero). That happens iff 60 is a cube mod p (and 6 is not).
The next few primes in your list are 151, 193, 199, 211, 223, 229, 277, 283, ...
So there must be some other trick to use for these p to find a
solution to the equation 3x^3+4y^3+5z^3=0 mod p, one which of
necessity has xyz nonzero. I don't recall what it is.
> However, the failure is somewhat finite in nature (analogous to how the class
> number provides a finite measure of the failure of unique factorization in the ring
> of integers of a > number field): define the curve C/Q : 3x3 + 4y3 + 5z3 = 0.
Um, I haven't read Mazur's paper but: (1) there is unique factorization in
the ring of integers, (2) the failure of unique factorization in the
rings of integers within other extensions of Q can indeed be "measured"
by the class group (or its cardinality, the class number), but (3) the
class group is not the same as the Selmer group: one is defined for
number fields, the other is defined for algebraic varieties. So, I don't know
where you are going here.
> It turns out that there exist exactly 5 curves (up to isomorphism over Q)
> which are isomorphic to C over each Qp. That is, globally isomorphic here is
> only "finitely" stronger than locally isomorphic.
Well, again there is a connection here but I think it's less explicit
than you are implying (explain). Isomorphisms between varieties can often be
described in terms of polynomial equations, so the _existence_ of an
isomorphism amounts to the existence of a point on a (different) variety,
and the nonexistence of such a rational point can sometimes by prime p :
no rational point can exist because no point can exist mod p. But the
converse does not hold (another example of the failure of local-to-global)
so you can indeed have two varieties that are isomorphic over each Q_p
and yet not be rationally isomorphic. But exactly what variety we're
talking about can be hard to describe.
Regarding your hint for a non-zero solution in Z, it is need true that the ring
of integers in Q( 6^{1/3}) is a free Z-lattice of rank 3. The integers which
have integral inverses are the units, and they are known to form a Z-lattice
of rank 1 with torsion subgroup of order 2 (because +1 and -1 are units).
So yes, there is an interesting group associated to this number field, and
the group is Z + Z/2 . But it's not the class group, and it's not the Selmer
group.
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| « Last Edit: Jan 25th, 2008, 9:10am by Michael Dagg » |
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Eigenray
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Re: 3x^3+4y^3+5z^3=0
« Reply #4 on: Jan 25th, 2008, 2:03pm » |
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on Jan 25th, 2008, 9:08am, Michael_Dagg wrote:| (1) there is unique factorization in the ring of integers, (2) the failure of unique factorization in the rings of integers within other extensions of Q can indeed be "measured" by the class group (or its cardinality, the class number), but |
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That's why I said "ring of integers of a number field".
Quote:(3) the class group is not the same as the Selmer group: one is defined for
number fields, the other is defined for algebraic varieties. So, I don't know
where you are going here. |
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It's an analogy. One can also view the class number as measuring a kind of local-to-global failure: locally, all ideals are principal, but not globally.
Quote:you can indeed have two varieties that are isomorphic over each Q_p
and yet not be rationally isomorphic. |
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Exactly. So global equivalence is a refinement of the partition into 'local equivalence' classes:
Any ideal of Z[ {-5}] (which all 'look like' (1) locally) looks globally like either (1) or (2,1+{-5}).
Any curve that looks like C locally looks globally like one of 5 different curves.
Just as the local equivalence class of an ideal breaks up into finitely many global equivalence classes, it is conjectured that the local equivalence class of a curve of genus 1 breaks up into finitely many global equivalence classes.
Quote:| So yes, there is an interesting group associated to this number field, and the group is Z + Z/2 . But it's not the class group, and it's not the Selmer group. |
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I never said it was! The code was only to show that the ring Z[61/3] is a UFD with fundamental unit  = -1 + 6*61/3 - 3*62/3. The proof is much more elementary if you assume these facts.
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Michael Dagg
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Re: 3x^3+4y^3+5z^3=0
« Reply #5 on: Jan 25th, 2008, 2:52pm » |
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Thanks for the clarification.
Lets try this: Suppose p>5 is prime. Let C = (Z/pZ)^*, and C^3 the
subgroup of cubes and consider the elements a=3/4, b=4/5, c=5/3 in C.
If any of a,b,c lies in C^3 then we can solve the equation
(*) 3 x^3 + 4 y^3 + 5 z^3 = 0
mod p by setting one variable to zero and another to 1.
If none of a,b,c lies in C^3 then note that since abc = -1 lies in C^3,
then the images of a,b,c in C/C^3 (a group of order 3) must be the
same: there's no other way to get three +-1's to add up to zero except
1+1+1 = (-1)+(-1)+(-1) = 0 mod 3. In particular, the quotient
a/b = 15/16 = 60/4^3 lies in C^3, i.e. 60 is a cube, say 60 = N^3.
Note that 3 N^3 + 4 (-5)^3 + 5 (4)^3 = 0 and so (*) is satisfied.
To complete the argument: note that the solution is solvable p-adically
because if we have an integer N making 3 N^3 + 4 (-5)^3 + 5 (4)^3 = 0
mod p^k , then 3 (N + t p^k)^3 + 4 (-5)^3 + 5 (4)^3 = 0 as long as
t = (60-N^3)/p^k * N * (180)^{-1} mod p. So we can successively compute
the terms in the p-adic expansion of N . A slightly different formula computes
a p-adic expansion for p=2 or 5, and for p=3 we can similarly compute a
p-adic solution to 3 0^3 + 4 y^3 + 5 = 0 : y = -2 + 9*1 + 27*1 + 81*1 + ... .
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| « Last Edit: Jan 25th, 2008, 2:55pm by Michael Dagg » |
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Eigenray
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Re: 3x^3+4y^3+5z^3=0
« Reply #6 on: Jan 25th, 2008, 7:54pm » |
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Ah, that's neat:
3*60 = 4*53 - 5*43,
4*60 = 3*53 - 5*33,
5*60 = 3*43 + 4*33.
Coincidence? 
What I had in mind was actually a bit different: if any of the cosets {3x3}, {4y3}, {5z3} are the same, we have a solution. Otherwise, they must exactly partition Zp*, so we have the solution (x, 1, -1), (1, -2y, 1), or (1, -1, z), depending on which of 3x3, 4y3, or 5z3, respectively, takes the value 1.
It turns out there's another argument which shows more generally that for any integers a,b,c,
ax3 + by3 + cz3 = 0
always has a non-trivial solution mod p (which can be lifted to Qp for all but finitely many p).
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Michael Dagg
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Re: 3x^3+4y^3+5z^3=0
« Reply #7 on: Jan 26th, 2008, 11:36am » |
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> Coincidence?
I'm reluctant to say "yes" because there may be a deep pattern here
that I hadn't noticed, but if it's not a coincidence, the greater
reason why these three equations all hold goes deep.
But I agree it's awfully strange that there are not one, not two, but
three equally good ways to work with the argument I gave, all of the
form 60 a = b c^3 +- c b^3 with {a,b,c} = {3,4,5} .
Kind of remarkable!
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| « Last Edit: Jan 26th, 2008, 11:36am by Michael Dagg » |
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balakrishnan
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Re: 3x^3+4y^3+5z^3=0
« Reply #8 on: Jan 27th, 2008, 3:44am » |
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on Jan 25th, 2008, 2:52pm, Michael_Dagg wrote:Thanks for the clarification.
Lets try this: Suppose p>5 is prime. Let C = (Z/pZ)^*, and C^3 the
subgroup of cubes and consider the elements a=3/4, b=4/5, c=5/3 in C.
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Can anyone enlighten what (Z/pZ)^* denotes.
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towr
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Re: 3x^3+4y^3+5z^3=0
« Reply #9 on: Jan 27th, 2008, 6:56am » |
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on Jan 27th, 2008, 3:44am, balakrishnan wrote:| Can anyone enlighten what (Z/pZ)^* denotes. |
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I think it's the multiplicative group modulo p (i.e. for prime p you have numbers 1..p-1, and you can multiply them modulo p; so for p=7, you have G={1,2,3,4,5,6}, and for any two elements g,h in G you have (g*h modulo 7)  G).
http://mathworld.wolfram.com/ModuloMultiplicationGroup.html
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| « Last Edit: Jan 27th, 2008, 7:00am by towr » |
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balakrishnan
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Re: 3x^3+4y^3+5z^3=0
« Reply #10 on: Jan 27th, 2008, 8:26am » |
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Thanks for the explanation,Towr.
It is still not very clear to me.
How does numbers like 3/4 (rational numbers) appear in the group?
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towr
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Re: 3x^3+4y^3+5z^3=0
« Reply #11 on: Jan 27th, 2008, 9:34am » |
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on Jan 27th, 2008, 8:26am, balakrishnan wrote:Thanks for the explanation,Towr.
It is still not very clear to me.
How does numbers like 3/4 (rational numbers) appear in the group? |
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Well, first, it's not entirely like regular maths. Any operation on elements from the group (using the group operator), gives another element of the group. So, in this case, every element is an integer in the range 1..p-1.
That doesn't mean 3/4 doesn't exist though. If 4 is an element of the group, so must 1/4 (it's a property of groups that the inverse of an element must be part of the group as well), and 3/4 = 3*1/4. So it exists; but it has to be equivalent to one of the integers in the group.
So, what can 1/4 be? It has to be the element such that when you multiply it by 4, you get 1 (modulo p). For p=7, we have 4*2 = 8  1 (mod 7), so 1/4 2 (mod 7).
This means 3/4 = 3*1/4 3*2 = 6 (and just to check, 6*4 = 24 3 (mod 7), so indeed multiplying it by 4 gives 3)
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| « Last Edit: Jan 27th, 2008, 9:41am by towr » |
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Icarus
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Re: 3x^3+4y^3+5z^3=0
« Reply #12 on: Jan 27th, 2008, 11:32am » |
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Z/pZ is the multiplicative group mod p, which is actually a field for prime p, as towr has shown.
The additional symbolism, ^*, is unknown to me, though, and I've been too lazy to try and track it down.
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I wonder: Which is larger
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Obob
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Re: 3x^3+4y^3+5z^3=0
« Reply #13 on: Jan 27th, 2008, 12:00pm » |
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Z/pZ is the field of integers mod p. For any ring R with unity, the symbol Rx or R* typically denotes the group of units of R under the operation of multiplication in the ring. So (Z/pZ)x and (Z/pZ)* both mean the multiplicative group of units of Z/pZ. Since Z/pZ is a field, this means that it is the multiplicative group of all nonzero elements of Z/pZ.
If by Z/pZ one means a group, then it should be the additive group of integers mod p, not the multiplicative group. The notation makes sense from the viewpoint of modern algebra because Z/pZ as a group is literally the integers modded out by the normal subgroup consisting of integers that are multiples of p. Likewise if by Z/pZ we mean a ring, then it is the ring Z modded out by the ideal of integers that are multiples of p. One gets the group Z/pZ by "forgetting" the multiplicative structure on the ring Z/pZ.
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Icarus
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Re: 3x^3+4y^3+5z^3=0
« Reply #14 on: Jan 27th, 2008, 12:04pm » |
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Heh. I thought the ^ meant something else. It didn't occur to me that it was only meant to represent superscripting.
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I wonder: Which is larger
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balakrishnan
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Re: 3x^3+4y^3+5z^3=0
« Reply #15 on: Jan 27th, 2008, 2:39pm » |
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Ah!
Thanks a lot Towr,Obor and Icarus.
Since p is a prime>5
3/4,4/5,5/3 are elements of the group.
Beautiful proof Michael 
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