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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

392.0. "Consec edges in tetrahedron" by TOOLS::STAN () Wed Nov 27 1985 22:06

Is there a tetrahedron whose 6 edges are consecutive integers
and which has integral volume?

I found the case of a tetrahedron with sides 6,7,8,9,10,11
having volume 48. (Base has sides 9, 10, 11; opposite sides
being 6, 7, 8, respectively.)

Are there any other examples?

T.R Title User Personal
Name Date Lines

392.1 R2ME2::GILBERT Sat Nov 30 1985 23:57 10

T.R	Title	User	Personal Name	Date	Lines
392.1		R2ME2::GILBERT		`Sat Nov 30 1985 23:57`	10
	Disregarding symettries (including reflections), there are 30 ways to label the edges of a tetrahedron with distinct values. Letting these values be n, n+1, n+2, ..., n+5, and using the equations of not 385, we get thirty diophantine equations. An integer solution to any of these will yield a tetrahedron with sides of consecutive integers, and having integral volume. Note that the form of these equations is: 288V^2 = p(n), where p(n) is a 6th degree integer polynomial in n. Note that to find other solutions, consideration of the equations in various modulii should prune the search space.

Disregarding symettries (including reflections), there are 30 ways to label
the edges of a tetrahedron with distinct values.  Letting these values be
n, n+1, n+2, ..., n+5, and using the equations of not 385, we get thirty
diophantine equations.  An integer solution to any of these will yield a
tetrahedron with sides of consecutive integers, and having integral volume.

Note that the form of these equations is:  288V^2 = p(n), where p(n) is a
6th degree integer polynomial in n.  Note that to find other solutions,
consideration of the equations in various modulii should prune the search
space.