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.

  • $\begingroup$ real solutions?integer solutions? $\endgroup$ – Avitus Sep 5 '13 at 14:49
  • $\begingroup$ Sorry… integer solutions! $\endgroup$ – Kieren MacMillan Sep 5 '13 at 15:08
  • $\begingroup$ To those who suggested the quadratic formula, a big thanks: that gives me enough to work with! I must admit, I feel a little silly (and humbled) not to have thought to use it myself. $\endgroup$ – Kieren MacMillan Sep 5 '13 at 15:30
  • $\begingroup$ But this will not help you, I think. You need something about the numbers which are representated by $x^2+3y^2$ $\endgroup$ – Dietrich Burde Sep 5 '13 at 15:36
  • $\begingroup$ Hmmmmm… that may well be. I'll report back after doing a little algebra. $\endgroup$ – Kieren MacMillan Sep 5 '13 at 15:46

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