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I am currently in the second year of high school doing a research paper in mathematics and I need some help with it.

I am trying to find numbers that are both Polygonal and Mersenne. Polygonal numbers are numbers of the form $\cfrac{n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides that the polygon has, a.k.a. an s-gon, and n is the number of points that one of the outermost sides has, a.k.a. the n-th s-gon. For example, when $n=2$ and $s=3$, we get the second triangular number, $3$. Mersenne numbers are numbers of the form $2^p -1$. The more known type of number is a Mersenne prime, but in this case we do not require the p to be a prime. So if we want to find numbers that are both Mersenne numbers and Polygonal numbers, we need to find integer solutions to the equation $\cfrac{n^2(s-2)-n(s-4)}{2} = 2^p -1$. First, in order to get rid of the redundant solutions, we say that $s \ge 3$ (because 3 is the smallest number of sides that a polygon can have (Euclidian space)) and that $n \ge 3$ (Because the first s-gonal number is always $1$, and the second s-gonal number is always $s$, with which we would be able to asign every Mersenne number atleast $1$ Polygonal number, but that wouldnt be fun :) ) and $p \ge 1$ ($p$ can be whatever positive integer).

So now we have an equation with 3 variables that we seek integer solutions to. I know that I will need to use Diophantines equations and others, but is it too hard for a mere mortal like me unlike the the mastermind that is Wolfram Alpha to solve this equation? Should I change it to instead of using the general formula for Polygonial numbers use the formula for a specific P. number (for example if we take triangular numbers we would get $\cfrac{n^2+n}{2} = 2^p -1$? What kind of questions should I ask after this one (which numbers are both P. and M.), maybe is there an infinite amount of them, which if I understand correctly is a typical question for Diophantine analysis.

Thanks in advance and please ask questions if you need further info., and sorry if there are some grammatical mistakes in the text.

Edit: I am seeking non-trivial solutions to the problem, as in n is not 1 or 2. Sorry about that, should have mentioned it earlier.

-Redelectrons

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Let's consider the approach of fixing either $n$ or $s$.

Fixing $s$ clearly leads to difficulties, for example for $s=3$, the problem becomes equivalent to the Ramanujan-Nagell equation whose solution is non-trivial - but interesting to understand/explore.

Fixing $n$ is probably a good approach, because the numbers now form an A.P. For example, at $n=3$, the numbers take the values of all multiples of 3 which are at least 6 and hence all $2^k-1$ values are taken for which $k$ is even and at least 4. You could get a collection of such "modulo" results for each value of $n$ and try to combine them.

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  • $\begingroup$ Thanks for the comment. What about not fixing any of the variables? $\endgroup$ – redelectrons Sep 29 '15 at 18:53

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