A conjecture about triangular numbers There is one and only one pair of natural numbers $m$ and $n$ ($15$ and $21$) such that: $m$ and $n$ are triangular numbers; the sum of their squares is a triangular number ($666$); and the sum of the triangular numbers $m(m+1)/2$ and $n(n+1)/2$ is a triangular number $k(k+1)/2$ ($k=26$). I have tested pairs of numbers from $m+n=2$ to $m+n=100 000 000$ using a computer program. (Notice that $15, 20, 25$ is a Pythagorean triple, giving the sides of a triangle that's similar to the sacred Egyptian triangle $3, 4, 5$.) I know, $666$ is the number of the beast and all that stuff, but this is a serious mathematical problem.
 A: You are looking for integral solutions to the system of equations
$$\left(\frac{a(a+1)}{2}\right)^2 + \left(\frac{b(b+1)}{2}\right)^2 = \frac{c(c+1)}{2}$$
$$\frac{1}{2}\left(\frac{a(a+1)}{2}\right)\left(\frac{a(a+1)}{2}+1\right) + \frac{1}{2}\left(\frac{b(b+1)}{2}\right)\left(\frac{b(b+1)}{2}+1\right) = \frac{d(d+1)}{2}$$
in the variables $a,b,c,d$. These equations describe an arithmetic surface $X$ in affine $4$-space over $\mathbf Q$. It can be viewed as a family of genus $2$ curves varying with the parameter $d$. Little is known about rational points on varieties of dimension $>1$. There are conjectures, for instance the Bomberi-Lang conjecture which is the higher-dimensional analogue of Falting's theorem. 
Such a problem is extremely difficult in general. It is quite possible that you have discovered the only solution. As for the significance of the number $666$, well, you'll have to ask a numerologist about that.
A: It may help to realize that the sum of the squares of two consecutive triangular numbers is always a triangular number.  That is,
$$
\left(\frac{1}{2}m(m+1)\right)^2+\left(\frac{1}{2}(m+1)(m+2)\right)^2=\frac{1}{2}(m+1)^2((m+1)^2+1)
$$
So, you could weaken the conjecture to 15 and 21 are the only consecutive triangular numbers $m$ and $n$ such that $m(m+1)/2+n(n+1)/2$ is a triangular number, and try to prove that.  
The 666 would then just be a bonus...
A: You know what else is crazy about this?
We see that even $(15 + 21 = 36)$ is also a triangular number so for the $n$th triangular number formula $\frac{n(n+1)}{2}$, we see that :-
$1$. $15$ is triangular with $n=5$, $21$ is triangular with $n=6$. So, as a previous comment noted, $15$ and $21$ are consecutive triangular numbers.
$2.$ $15^2 + 21^2 = 666$ is triangular with $n=36$ .
$3.$ $\frac{15(15+1)}{2}+\frac{21(21+1)}{2} = 351$ is triangular with $n=26$.
$4.$ $15+21=36$ is triangular with $n=8$ (actually, the sum of any two consecutive triangular numbers is always a perfect square. what’s up with a triangular square number showing up here?)
What are the connections between these values of $n$? What makes $15$ and $21$ ($n=5$ and $n=6$) so special? And of course $666$ is thrown into the mix here too.
Does it mean something deeper in regards to our universe? Probably not. But this is really interesting.
