# Pentagonal-Triangular numbers

Pentagonal Triangular Number is a number which is simultaneously a pentagonal number $P_n$ and triangular number $T_m$ . Such numbers exist when $$\frac{1}{2}n(3n-1) = \frac{1}{2}m(m+1)$$ This problem boils down to the pell-like quadratic Diophantine equation: $$x^2-3y^2 = -2$$ Using Brahmagupta's Lemma: $$\text{Theorem 2.22 (Brahmagumpta)}\;\; \text{If }\; u^2 − Pv2 = k \;\; \text{ and } \;\; r^2 − Ps^2 = l \;\;\;\text{then} (ur ± Pvs)^2 − P(us ± vr)^2 = kl.$$ One can prove that there are infinitely many $(n,m)$ such that $P_n = T_m$

(I did this around $3$ years back for extra credit in my number theory class)

I was just wondering if there were already papers or instances where somebody already showed this.
Thanks.

### EDIT

I explicitly ask for papers here and not websites where they discuss the general topic. I'm looking for a formal proof of something I've already witnessed so that I may compare their methodology with mine.

• Jan 27, 2013 at 22:39
• According to this page, there are recursive formulas for the $n$th Pentogonal-Triangular number. I do not see an upper limit on these formulas, so I assume this implies that there are infinitely many such numbers, right? oeis.org/A046174 Jan 27, 2013 at 22:45
• @anorton Hey, I'm sorry. Thank you. It's not you, (Several people have just pointed to that site when I've already seen it). I appreciate your desire to help me, thank you again. Jan 27, 2013 at 23:07
• One could just define sequences of integers $x_k$ and $y_k$ by $$x_k + y_k \sqrt{3} = (1+\sqrt{3}) (2 + \sqrt{3})^k$$ and note that $x_k^2-3y_k^2=-2$. Jan 28, 2013 at 0:07
• Ristyn, if you don't want people to tell you about webpages and things you already know about, the thing to do is to include that information when you first post the question. That also stops others from wasting their time finding things you already know about. That said, have you followed all the links on the wolfram and oeis pages to see whether any of them point to proofs? And that said, even if I can't point to a page where it was explicitly done, solving $x^2-3y^2=-2$ is standard stuff, I'm sure it has been done. Jan 28, 2013 at 0:10