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Solving multilinear integer equations

asked 2012-02-06 02:06:08 -0500

updated 2012-02-08 04:27:35 -0500

I'd like to explore the solutions of a multilinear diophantine equation like $12(yz) + 6(y+z) +2 -4xw -2(w+x) = 0$. In particular I'd like to generate instances of solutions. I tried the function solve like:

(x,y,z,w) = var('x,y,z,w')
assume(x, 'integer'); assume(y, 'integer')
assume(z, 'integer'); assume(w, 'integer')
solve([12*(y*z) + 6*(y+z) +2 -4*x*w -2*(w+x) == 0, y==y], x)

which gives me the not particularly useful output:

[[x == (3*(2*r1 + 1)*r2 + 3*r1 - r3 + 1)/(2*r3 + 1), y == r1, z == r2, w == r3]]

I'd like to generate some actual quadruples that solve this equation or possibly investigate the structure more. Is there some systematic way to do this? In particular, when playing around with coefficients, is there a test of existence of an integer solution?

Edit: I had typos in the first version, I edited the equation.

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Is this helpful?

sage: maxima('solve(12(yz) + 6(y+x) +2 -4xy -2(w+x) = 0, [x,y,z,w])').sage()

[[x == r1, y == r2, z == r3, w == -(2r1 - 3)r2 + 6r2r3 + 2*r1 + 1]]

achrzesz ( 2012-02-06 04:51:41 -0500 )edit

Hmm, but those are real solutions, not integer, though in this case it looks like it will work if you use all integers.

kcrisman ( 2012-02-06 07:19:02 -0500 )edit

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answered 2012-02-06 07:29:24 -0500

kcrisman gravatar image

This is NOT a 'real' answer - achrzesz had the real idea. Maybe he should post it for credit. This is just it in Sage.

sage: (x,y,z,w) = var('x,y,z,w')
sage: solve([12*(y*z) + 6*(y+x) +2 -4*x*y -2*(w+x) == 0,y==y], x,y,z,w)
[[x == r6, y == r7, z == r8, w == -(2*r6 - 3)*r7 + 6*r7*r8 + 2*r6 + 1]]

You may ask why we need the dummy equation y==y. See this Trac ticket.

This does not answer the question of how to treat this as a non-linear Diophantine equation. I seem to recall that this is undecidable, but maybe I'm misinterpreting your question. You can always just do something really naive like loop through all integers. I don't know that we have anything more efficient that that for integers.

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answered 2012-02-06 11:46:07 -0500

Volker Braun gravatar image

You can divide by $2$ and solve for $w$. So the set of all solutions is parametrized by $(x,y,z)\in\mathbb{Z}^3$ arbitrary.

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Well, Thomas did say equations like that one, not just that one...

kcrisman ( 2012-02-06 13:40:03 -0500 )edit

Well, I have a typo even in the first example :( My starting point was $4nyz + 2n(y + z) + n - 4wx - 2(w + x) ? 1 = 0$ for odd $n$.

Thomas ( 2012-02-08 04:21:06 -0500 )edit

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Asked: 2012-02-06 02:06:08 -0500

Seen: 704 times

Last updated: Feb 08 '12