I want to prove -- using elementary math only -- that the following equation has no integer solutions for $t \ge 1$:
$$ 6a^2(16a^2+1) = \frac{t(t+1)}{2}. \qquad(1)$$
I know it doesn't (or at least shouldn't), because I derived it from a known equation with only one solution ($t=a=0$). Just to be sure, I also confirmed (using maxima) that there are no solutions with $1 \le a \le 10000$.
Any thoughts on how to continue algebraically from $(1)$ would be appreciated.
EDIT: Another equivalent reduction (if it's easier to work with) would be
$$ (6x^2+1)^2+3(2x^2)^2 = y^2. $$
[Proof: Clearly y is odd. Now reducing modulo 8, we see that x must be even. Substitute y=2t+1 and x=2a and simplify to get (1).]
Thanks! Kieren.