all(?) solutions
It just struck me that solving 2R^2=T^2+1 is related to the continued fraction for sqrt(2).
The bottom line is that if (R,T) is a solution, then so is (3R+2T,4R+3T).
Starting with (R,T)=(1,1) you generate infinitely many solutions, and I think these might be all.
So Solutions are (R,T)=(1,1), (5,7), (29,41), (169,239), ...
Then translate back to get solutions (r,t) to t(t-1) = 2r(r-1).
|