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Mark Spahn science forum addict
Joined: 07 Jul 2005
Posts: 62

Posted: Mon Jun 26, 2006 11:25 pm Post subject:
Nicomachus's theorem



Nicomachus's theorem says
1^3 + 2^3 + ... + n^3 = (1 + 2 + ... +n)^2
(I learned its name at http://mathworld.wolfram.com/PowerSum.html).
Does this equality have any geometric interpretation
(probably not; the left side is a volume, the right side is an area)?
I think I ran across an explanation of a kind of cryptography
based on this equality. At the time I was in a hurry and I
skipped over it. Or perhaps I am simply misremembering what
I saw. At any rate, I cannot find it anymore.
Does anyone know of an application of Nicomachus's theorem
to cryptography?
 Mark Spahn 

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mensanator@aol.compost science forum Guru
Joined: 24 Mar 2005
Posts: 826

Posted: Tue Jun 27, 2006 1:03 am Post subject:
Re: Nicomachus's theorem



Mark Spahn wrote:
Yep.
Quote:  (probably not; the left side is a volume, the right side is an area)?

But a volume can be seen as 3 areas stacked on top of each other.
See
<http://members.aol.com/mensanator/sumofcubes.htm>
Quote:  I think I ran across an explanation of a kind of cryptography
based on this equality. At the time I was in a hurry and I
skipped over it. Or perhaps I am simply misremembering what
I saw. At any rate, I cannot find it anymore.
Does anyone know of an application of Nicomachus's theorem
to cryptography?
 Mark Spahn 


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Fred Curtis science forum beginner
Joined: 26 Aug 2005
Posts: 2

Posted: Tue Jun 27, 2006 1:05 am Post subject:
Re: Nicomachus's theorem



Mark Spahn <mspahn@localnet.com> wrote:
Quote:  Nicomachus's theorem says
1^3 + 2^3 + ... + n^3 = (1 + 2 + ... +n)^2
(I learned its name at http://mathworld.wolfram.com/PowerSum.html).
Does this equality have any geometric interpretation
(probably not; the left side is a volume, the right side is an area)?

Here's one geometric interpretation: The RHS of the identity is
a square arrangement, (1+2+...+n) on each side, of unit cubes.
The (geometric) cubes of the LHS can be cut into kxkx1 slices to
occupy the same volume.
Here is an arrangement of cubeslices for n=5. For all k, there
are k kxkx1 cube slices; for even k, one of the slices is broken
into two (k/2)xkx1 halves.
1.....444455555
..22...444455555
..22.......55555
....333....55555
....333....55555
....333.........
44....4444.....
44....4444.....
44....4444.....
44....4444.....
55555.....55555
55555.....55555
55555.....55555
55555.....55555
55555.....55555
55555.....55555
To see inductively that the cube slices fit, observe that for k,
the kxkx1 slices are fitted around a square region with sidelength
(1+2+...+k1) = (k1)k/2, so the number of kxkx1 slices that will
surround this region are (k1)/2 on each of the two sides, plus
the slice touching the region's corner, for a total of k slices. 

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Patrick Hamlyn science forum beginner
Joined: 03 May 2005
Posts: 45

Posted: Tue Jun 27, 2006 2:27 am Post subject:
Re: Nicomachus's theorem



"Mark Spahn" <mspahn@localnet.com> wrote:
Quote:  Nicomachus's theorem says
1^3 + 2^3 + ... + n^3 = (1 + 2 + ... +n)^2
(I learned its name at http://mathworld.wolfram.com/PowerSum.html).
Does this equality have any geometric interpretation
(probably not; the left side is a volume, the right side is an area)?
I think I ran across an explanation of a kind of cryptography
based on this equality. At the time I was in a hurry and I
skipped over it. Or perhaps I am simply misremembering what
I saw. At any rate, I cannot find it anymore.
Does anyone know of an application of Nicomachus's theorem
to cryptography?
 Mark Spahn

Can't help with cryptography, but there's a nice geometric interpretation  look
up Partridge Tilings.
Basically:
Take one unit square, two squares of side 2, three squares of side 3,... up to n
squares of side n (these have a total area of 1^3 + 2^3 + 3^3 + ... + n^3)
Now use them to tile a square of side (1+2+3...+n).
Repeat with any nonsquare shapes you can think of.
The Partridge comes from the pear tree in the Christmas song of course. Bob
Wainwright came up with the idea and the name, and presented it at Gathering for
Gardner 2.
http://www.stetson.edu/~efriedma/mathmagic/0802.html
http://www.mathpuzzle.com/partridge.html 

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