Diophantus: Dato numero, invenire quot modis multangulus esse possit.

[Acme]

Dato numero, invenire quot modis multangulus esse possit, or Given a number, to find in how many ways it can be polygonal.

 

I am looking for historical solutions to this problem of Diophantus, whether old or recent. I have found that Fermat implied he had a solution, but he gave no actual solution. (as if. lol) Any help is appreciated. Thanks.

[DrRocket]

Dato numero, invenire quot modis multangulus esse possit, or Given a number, to find in how many ways it can be polygonal.

 

I am looking for historical solutions to this problem of Diophantus, whether old or recent. I have found that Fermat implied he had a solution, but he gave no actual solution. (as if. lol) Any help is appreciated. Thanks.

 

Fundamentals of Diophantine Geometry by Serge Lang would be a good place to start.

[Acme]

Fundamentals of Diophantine Geometry by Serge Lang would be a good place to start.

 

Thanks. I see the volume goes for $92.00 as an e-book so it could be some time before I could purchase it as I am retired & on a small fixed income. Google books gives only a few excerpts and they do not address my question. I doubt my local community library has it but I will check. My community has no universities either, though we do have a community college which I can check too.

 

Do you know for a fact the solution(s) I seek are in the book? Do you have the book and if so are you willing to check it for me? Do you know of a solution or if one exists?

 

So as not to appear too "dodgy" I should say that I do know a solution though it is not mine. It can be written in a short paragraph and does not involve any complex mathematics at all. What I want to get at is if there are any other extant solutions and if so who wrote them and when and how they compare to the solution I have.

 

You [all] of course understand my reticence to believe Fermat's claim and I beg everyone's forebearance in not giving the solution I have for the time being. In part I want to avoid presenting a bias and while I do have permission to use it I would want to touch base with the author again before posting it here. (Assuming anyone is interested in such a thing.)

 

 

Thanks again.

[DrRocket]

Proofs are known.

 

http://www.archive.org/details/diophantusofalex00heatiala

 

http://en.wikipedia.org/wiki/Fermat_polygonal_number_theorem#CITEREFHeath1910

[Acme]

Proofs are known.

 

http://www.archive.o...falex00heatiala

 

http://en.wikipedia....ITEREFHeath1910

 

 

Excellent!! Thank you!! This has been a long-standing question for me/us. Here is my friend's version of a solution and another fella's variation for comparison. I will pass this all on to them as they came about the solutions independently. Coinikydinkily, the variant example I'm giving uses the same number 325 as the example on page 259 in the track you gave. I will be looking at the geometric analysis a bit more as we came at things only algebraically from the generalized expression for polygonals.

 

fella 1: I'd handle the problem this way:

 

F = 1/2((n^2)*s) - 2*n^2 - n*s + 4*n)

147 = 1/2((n^2)*s) - 2*n^2 - n*s + 4*n)

294 = ((n^2)*s) - 2*n^2 - n*s + 4*n)

 

294 factorises as 1,2,3,7,7. One of those factors must be n

 

checking n=1: 294 = s - 2 - s + 4 = 2: Clearly n=1 doesn't work

checking n=2: 294 = 4s - 8 - 2s + 8 =2s: Gives s=147. A trivial result: n=2 goes in steps of 1, generating

 

every number.

checking n=3: 294 = 9s - 18 - 3s + 12 = 6s - 6: Gives 6s=300, s=50

checking n=4: 294 = 49s - 98 -7s + 28 = 4ss - 70: Gives 42s=364, =8.6666666666666666666666666666667 which is

 

fractional and therefore not a solution.

 

answer: n=3, s=50.

 

fella 2: alternate method given

 

factors of 325*2

650 2, 5, 10, 13, 25, 26, 50, 65, 130, 325

 

650/5 = 130 -2 = 128/4 = 32

 

650/10 = 65 -2 = 63/9 = 7 + 2 = 9

 

650/13 = 50 -2 = 48/12 = 4 + 2 = 6

 

650/25 = 26 -2 = 24/24 = 1 + 2 = 3

 

basically, if a number, multiplied by 2 divides n, and then after division by n and subtracting 2, is

 

divisible by n-1, then it's figurate. in short 2*F = n*((n-1)*a +2)

which gon it is is determined after dividing by n-1. just add two the result, with n being the term.

 

Fella 2 has also been kind enough to do some programming for me and set me up with portable Python so that I could generate sequential extensive lists of the polygonals which I have done on the interval 6 to 1000000000. In that interval I found the greatest "multiplicity" to be 17 at the number 879207616 with the following solutions [n,s]. [4, 146534604, 7, 41867031, 14, 9661624, 16, 7326732, 28, 2325948, 31, 1890771, 56, 570916, 58, 531888, 118, 127368, 236, 31708, 248, 28708, 406, 10696, 496, 7164, 518, 6568, 1711, 603, 2146, 384, 7192, 36]

 

This is all rather incidental to my interest in the set of numbers which are not polygonal and perhaps I could impose on you further for any information/knowledge you might have of that set. ? The OEIS listing of the set gives credit for the set listing to a fella named Beiler but Beiler's cited work we have checked and found it does not give that set at all. (OEIS actually gives the set incorrectly in listing {1,2,3,4,5} as members in spite of the restrictions n & s >=3.)

 

Anyway, this is a recreational pursuit of mine that keeps me off the streets and I sincerely appreciate the help and interest. If you do have an interest in the non-polygonal set, I have independent work o'plenty to share. Thanks again.

[Acme]

Just dropping in to drop off some eye-candy. The design is mine, the execution is computer generated by fella_2 and recolored by hand by moi. Not sure if spiral arrays are original to me, but I came up with the idea independently. The array begins in the middle, spirals out clockwise, and sequential cells represent the natural numbers. In this particular view the black cells are polygonal numbers and white not. Thanks for having a taste & hoping it's as fun to eat as it was to cook.

[Acme]

Noting again the OEIS error in listing {1,2,3,4,5} with the set of non-polygonal numbers, I want to point out that while 3 is technically the second triangular number (and 1 is the first), it is proper to exclude them and all n= 1 or 2 values for all polygonals of side s. Otherwise, all numbers would be trivially polygonal. OEIS does properly show this restriction but does not follow it with the listing. Just so, and following the restriction, it can be proved that all multiples of 3 are polygonal and so no non-polygonal numbers are multiples of 3. Here's my proof.

 

 

[math]F=frac{1}{2}(n^2s-2n^2-ns+4n)[/MATH]

 

let n = 3

 

let s = {2,3,4,5...} set of integers >=2

 

[math]F=frac{1}{2}(3^2s-2*3^2-3s+4*3)[/MATH]

 

[math]F=frac{1}{2}(9s-18-3s+12)[/MATH]

 

[math]F=frac{1}{2}(6s-6)[/MATH]

 

[math]F=(3s-3)[/MATH]

 

[math]F=3(s-1)[/MATH]

 

Δ, all multiples of 3 are figurate numbers

 

Δ, no non-figurate numbers are multiples of 3

 

(sorry for the latex errors; little help?)

Edited March 2, 2012 by Acme