The following conjecture is about polygonal number sequences:
For every $r\ge{3},s\ge{3} \in \mathbb{N}$, there exist $m\ge{2},n\ge{2} \in \mathbb{N}$ such that $$\frac{n^{2}(r-2)-n(r-4)}{2}\\=\frac{m^{2}(s-2)-m(s-4)}{2}$$
In other words, all sequences of polygonal numbers share some of their terms with each of the others, but not the same ones, obviously. This has already been shown in specific cases such as 36, which is both a square and a triangular number:
$$36=\frac{8(9)}{2}=6^2$$
The conjecture above appears to hold true in every case in which it is tested, but how would I prove it? Has it already been proven? If it's not true for some case, how would I disprove it? I am looking for a non-trivial proof.
The conjecture is false.
Take $(r,s)=(14,5)$. Suppose that there exist $m\ge 2,n\ge 2\in\mathbb N$ such that
$$3m^2-m-12n^2+10n=0$$ which implies $$m=\frac{1\pm\sqrt{144n^2-120n+1}}{6}$$
There has to exist a non-negative integer $t$ such that $$144n^2-120n+1=t^2$$ which implies $$\begin{align}&(12n-5)^2-t^2=24 \\\\&\implies (12n-5-t)(12n-5+t)=24 \\\\&\implies (12n-5-t,12n-5+t)=(2,12),(4,6),(-12,-2),(-6,-4) \\\\&\implies (n,t)=(1,5),\left(\frac 56,1\right),\left(-\frac 16,5\right),(0,1)\end{align}$$ which contradicts that $n\ge 2\in\mathbb N$.
Added : There are many counterexamples.
Take $(r,s)=(16p+6,4p+3)$ where $p\ge 5$ is a Sophie Germain prime.
Suppose that there exist $m\ge 2,n\ge 2\in\mathbb N$ such that $$(4p+1)m^2-(4p-1)m+(-16p-4)n^2+(16p+2)n=0$$ which implies $$m=\frac{4p-1\pm\sqrt{16(4p+1)^2n^2+8(4p+1)(-8p-1)n+(4p-1)^2}}{2(4p+1)}$$
There has to exist a non-negative integer $t$ such that $$16(4p+1)^2n^2+8(4p+1)(-8p-1)n+(4p-1)^2=t^2$$ which can be written as $$((16p+4)n-8p-1+t)((16p+4)n-8p-1-t)=24p(2p+1)$$ Since both $p$ and $2p+1$ are odd primes, we get $$((16p+4)n-8p-1+t,(16p+4)n-8p-1-t)$$ $$=(12p(2p+1),2),(6p(2p+1),4), (4p(2p+1),6),(2p(2p+1),12),$$ $$(12(2p+1),2p),(6(2p+1),4p),(12p,2(2p+1)),(4(2p+1),6p)$$ implying $$n=\frac{6p^2+7p+1}{4p+2},\frac{6p^2+11p+3}{16p+4},\frac{2p^2+5p+2}{8p+2},$$ $$\frac{2p^2+9p+7}{16p+4},\frac{21p+7}{16p+4},1,\frac{8p+1}{8p+2},\frac{15p+3}{16p+4}$$ Considering $16n$, we have $$16n=\frac{48p^2+56p+8}{2p+1},\frac{24p^2+44p+12}{4p+1},\frac{16p^2+40p+16}{4p+1},$$ $$\frac{8p^2+36p+28}{4p+1},\frac{84p+28}{4p+1},16,\frac{128p+16}{8p+2},\frac{240p+48}{16p+4},$$ i.e. $$16n=24p+16-\frac{8}{2p+1},6p+9+\frac{2p+3}{4p+1},4p+9+\frac{7}{4p+1},$$ $$2p+8+\frac{2p+20}{4p+1},21+\frac{7}{4p+1},16,16-\frac{8}{4p+1},15-\frac{3}{4p+1}$$ This contradicts that $n\ge 2\in\mathbb N$.