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I posted the same question on Math SE since this one got put on hold.
Link to Math SE question:Polygonal Mersenne numbers

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get $$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$ where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about polygons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ 1 $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreciated. Also please ask if you require further details.
(Is it still too vague what I am asking for?)

[1] Greg Martin
[2] user236182

-redelectrons

I posted the same question on Math SE since this one got put on hold.
Link to Math SE question:Polygonal Mersenne numbers

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get $$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$ where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about polygons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ 1 $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreciated. Also please ask if you require further details.
(Is it still too vague what I am asking for?)

[1] Greg Martin
[2] user236182

-redelectrons

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get $$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$ where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about polygons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ [1] $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreciated. Also please ask if you require further details.
(Is it still too vague what I am asking for?)

[1] Greg Martin
[2] user236182

-redelectrons

I posted the same question on Math SE since this one got put on hold.
Link to Math SE question:Polygonal Mersenne numbers

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get $$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$ where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about polygons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ 1 $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreciated. Also please ask if you require further details.
(Is it still too vague what I am asking for?)

[1] Greg Martin
[2] user236182

-redelectrons

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get $$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$ where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about polygons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ [1] $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreaciated. Also please ask if you require further details.
(Is it still too vague what I am asking for?)

[1] Greg Martin
[2] user236182

-redelectrons

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get $$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$ where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about polygons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ [1] $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreciated. Also please ask if you require further details.
(Is it still too vague what I am asking for?)

[1] Greg Martin
[2] user236182

-redelectrons