[math-fun] Solving polynomial diophantine equations
This question is called "Hasse / local-global principle" and there is a large literature on its truth and falsity in various situations. For example, re Goucher's 4 squares example, relevant is "Hasse's theorem": Consider F(x1,x2,...,xm)=0 with F a homogeneous quadratic polynomial. This has a solution over Z^m iff it has a solution over Q^m, iff it both has a solution mod N for every N>1 and has a real-number solution, iff it both has a solution mod P^K for every K>0 and prime P and has a real-number solution. However, the Hasse principle fails for homogeneous cubics. Example by Selmer 3*X^3 + 4*Y^3 + 5*Z^3 = 0 has no nontrivial integer solutions despite having both real solutions and solutions modulo N for every N>1. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Asimov: It's known that the equation 1^2 + . . . + n^2 = m^2 has only one non-trivial solution: (n,m) = (24,70). 4) What solutions (n,m) exist if the exponent 2 is replaced everywhere by an integer p > 2 ??? --the way uniqueness was shown for exponent p=2 was (a) computer finds all small solutions, (b) this problem is an "elliptic curve" which known technology will resolve. The case p=3 should be settleable by the same method since it too yields an elliptic curve? Or is this not elliptic? Anyhow, in general Faltings' theorem, solving the Mordell conjecture, should be applicable to show that all of these problems (for each p) have only a finite number of solutions each (at most)? I have not actually gone thru the details so caveat emptor -- I'm merely trying to point you in right directions. If you can find a modulus N under which some such equation is not soluble (easy to find such N using computer if exist) then of course no integer solutions can exist.
The p=3 case is indeed an elliptic curve, by a special hack: The equation becomes [n(n+1)/2]^2 = m^3, so m must be a square, say m = k^2; then n(n+1)/2 = k^3, which is an elliptic curve. Pushing a tad further gives 4n^2 + 4n + 1 = (2n+1)^2 = (2k)^3 + 1. The only solutions are (+-1)^2 = 0^3+1 and (+-3)^2 = 2^3+1, which lead only to trivial solutions of the original problem. Rich PS: When doing number theory, try to reserve "p" for primes. It saves trouble for those of us who skip over details. --R ----- Quoting Warren D Smith <warren.wds@gmail.com>:
Asimov: It's known that the equation 1^2 + . . . + n^2 = m^2 has only one non-trivial solution: (n,m) = (24,70).
4) What solutions (n,m) exist if the exponent 2 is replaced everywhere by an integer p > 2 ???
--the way uniqueness was shown for exponent p=2 was (a) computer finds all small solutions, (b) this problem is an "elliptic curve" which known technology will resolve.
The case p=3 should be settleable by the same method since it too yields an elliptic curve? Or is this not elliptic? Anyhow, in general Faltings' theorem, solving the Mordell conjecture, should be applicable to show that all of these problems (for each p) have only a finite number of solutions each (at most)?
I have not actually gone thru the details so caveat emptor -- I'm merely trying to point you in right directions. If you can find a modulus N under which some such equation is not soluble (easy to find such N using computer if exist) then of course no integer solutions can exist.
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Sorry about that. I actually agonized for a couple of seconds over what letter to use. Thought first of e for exponent, but rejected it since e = 2.7 . . .. So what *is* a good letter for exponent? It's easy to see why they resorted to Greek letters. --Dan On 2013-08-28, at 2:59 PM, rcs@xmission.com wrote:
PS: When doing number theory, try to reserve "p" for primes. It saves trouble for those of us who skip over details. --R
You've rediscovered Euler's sum of powers conjecture: http://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture Victor On Tue, Aug 27, 2013 at 8:18 PM, Warren D Smith <warren.wds@gmail.com>wrote:
Asimov: It's known that the equation 1^2 + . . . + n^2 = m^2 has only one non-trivial solution: (n,m) = (24,70).
4) What solutions (n,m) exist if the exponent 2 is replaced everywhere by an integer p > 2 ???
--the way uniqueness was shown for exponent p=2 was (a) computer finds all small solutions, (b) this problem is an "elliptic curve" which known technology will resolve.
The case p=3 should be settleable by the same method since it too yields an elliptic curve? Or is this not elliptic? Anyhow, in general Faltings' theorem, solving the Mordell conjecture, should be applicable to show that all of these problems (for each p) have only a finite number of solutions each (at most)?
I have not actually gone thru the details so caveat emptor -- I'm merely trying to point you in right directions. If you can find a modulus N under which some such equation is not soluble (easy to find such N using computer if exist) then of course no integer solutions can exist.
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As you may know, the complex projective plane (CP^2) is defined as the equivalence classes of C^3 - {(0,0,0)} under the equivalence relation (x,y,z) ~ (cx,cy,cz) for any nonzero complex number c. Which is the same as letting S^5 = {(x,y,z) in C^3 | |x|^2 + |y|^2 + |z|^2 = 1}, factored out by the equivalence relation (x,y,z) = (ux,uy,uz), where u is any complex number with |u| = 1. Each equivalence class is clearly a great circle of S^5. So, CP^2 is a 4-manifold. (Some people like to think of it as a complex 2-manifold, but be that as it may.) The standard metric on CP^2 is defined from this last version, with the distance between points of CP^2 being defined by the distance between the corresponding great circles of S^5. (Every point of either circle has the same distance to the other circle -- through S^5 -- as any other point of it does.) It's easy to see that this geometry must be homogeneous: for any 2 points p, q of CP^2, there is an isometry of CP^2 taking p to q. (It's not isotropic, though.) With respect to this standard metric, it's interesting to ask what happens as you get farther and farther from a given point p of CP^2. Turns out that the locus of all points of CP^2 at a distance D from p form a 3-sphere S^3, for 0 < D < pi/2. As D increases towards pi/2, the size of this S^3 increases toward a supremum of radius 1. But the supremum is never reached. Because suddenly at D = pi/2, the locus changes from a 3-sphere to a 2-sphere (of radius = 1/2). And pi/2 is the maximum distance in CP^2 (i.e., its diameter), so you can't get any farther from p. And this must be true no matter which p we started with. This seems to be almost paradoxical behavior for a manifold that is homogeneous.
You might find it interesting to think of CP^2 as a "qutrit": rank one 3x3 Hermitian matrices with unit trace. At least now you have another search term! Example: http://arxiv.org/abs/1111.4427. Physicists often relax the rank one property, in which case the qutrit is in a "mixed-state" (as opposed to a "pure state"). -Veit On Aug 28, 2013, at 12:02 AM, Dan Asimov <dasimov@earthlink.net> wrote:
As you may know, the complex projective plane (CP^2) is defined as the equivalence classes of C^3 - {(0,0,0)} under the equivalence relation (x,y,z) ~ (cx,cy,cz) for any nonzero complex number c.
Which is the same as letting S^5 = {(x,y,z) in C^3 | |x|^2 + |y|^2 + |z|^2 = 1}, factored out by the equivalence relation (x,y,z) = (ux,uy,uz), where u is any complex number with |u| = 1. Each equivalence class is clearly a great circle of S^5.
So, CP^2 is a 4-manifold. (Some people like to think of it as a complex 2-manifold, but be that as it may.)
The standard metric on CP^2 is defined from this last version, with the distance between points of CP^2 being defined by the distance between the corresponding great circles of S^5. (Every point of either circle has the same distance to the other circle -- through S^5 -- as any other point of it does.)
It's easy to see that this geometry must be homogeneous: for any 2 points p, q of CP^2, there is an isometry of CP^2 taking p to q. (It's not isotropic, though.)
With respect to this standard metric, it's interesting to ask what happens as you get farther and farther from a given point p of CP^2. Turns out that the locus of all points of CP^2 at a distance D from p form a 3-sphere S^3, for 0 < D < pi/2. As D increases towards pi/2, the size of this S^3 increases toward a supremum of radius 1.
But the supremum is never reached. Because suddenly at D = pi/2, the locus changes from a 3-sphere to a 2-sphere (of radius = 1/2). And pi/2 is the maximum distance in CP^2 (i.e., its diameter), so you can't get any farther from p. And this must be true no matter which p we started with.
This seems to be almost paradoxical behavior for a manifold that is homogeneous.
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participants (5)
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Dan Asimov -
rcs＠xmission.com -
Veit Elser -
Victor Miller -
Warren D Smith