# Hexagonal Triangular Squares

Is there a hexagonal, triangular, square (apart from 0 and 1)?
In other words, is there a positive integer that is simultaneously

(1) a perfect square, $n^2$, $n \ge 2$,

(2) a triangular number, $\frac{m(m+1)}{2}$, $m$ an integer,

and (3) a (centered) hexagonal number, $(p+1)^3 - p^3$, $p$ an integer?

• I'll repeat the comment I've made on this problem before: "I’m way to lazy to work out the details right now, but: the equations you are dealing with are a pair of quadratic equations in three variables. That means they probably cut out an elliptic curve. Assuming this is correct, we know from Siegel’s Theorem that there are only finitely many integral points." topologicalmusings.wordpress.com/2008/09/28/… – David E Speyer Jul 14 '10 at 22:36

The question is equivalent to the system of quadratic Diophantine equations: $$8 n^2 = m'^2 - 1$$ $$4n^2 = 3p'^2 + 1$$ where $m'=2m+1$ and $p'=2p+1$. How to solve such systems is described in my paper: http://arxiv.org/abs/1002.1679 (see Theorem 6)
It is easy to obtain that the only nonegative solution is $n=1$, $m=1$, $p=0$.