Show that there are infinitely many solutions $(t, d)$ to $c - d^2 = t^2$ where $2c-1 = 3t^2$.

Question

Set $c = (3t^2+1)/2$.

Prove that infinitely many times, $c - t^2 = d^2$ for positive integers $t, d$. The first such instance is with $t = 1, d = 1$, and $t = 7, d = 5$. Can anyone come up with more solutions. Thanks for help.

$2 = (3\cdot1^2+1)/2$, and $2-1^2=1^2$

$74 = (3\cdot7^2+1)/2$, and $74-7^2=5^2$

Answer 1

This is $$\frac{3t^2+1}2-d^2=t^2$$ that is $$t^2-2d^2=-1.$$ This is a negative Pell equation. Its positive solutions are given by $$t+d\sqrt2=(1+\sqrt2)^m$$ where $m$ is an odd positive integer.