3 $"On a Diophantine Equation" paper of Erdös, at some point it is said that it is well known that$C(n,2)=x^2$has infinitely many integer solutions. I am just wondering the formula generating all possible$n \geq 2$,because I wonder whether they all come by integer solutions of$2x^2-y^2=\pm 1$Pell's equation generated by fundamental unit, or is it possible to have different$n$'s that do not come from the solutions of the pell's equation. For example$C(2,2)=1$a square,$C(9,2)=36$a square too, so the first two$n$'s are 2,9. One can also see from the solutions of$2x^2-y^2=\pm 1$pell's equation, by take$x=1, y=0$for$n=2=2*1$and take$x=2, $y=3$ for $n=9=3^2$.
"On a Diophantine Equation" paper of Erdös, at some point it is said that it is well known that C(n,2)=x^2 $C(n,2)=x^2$ has infinitely many integer solutions. I am just wondering the formula generating all possible n>=2,because $n \geq 2$,because I wonder whether they all come by integer solutions of 2x^2-y^2=-+1 pell's $2x^2-y^2=\pm 1$ Pell's equation generated by fundamental unit, or is it possible to have different n's $n$'s that does do not come from the solutions of the pell's equation.
For example C(2,2)=1 $C(2,2)=1$ a square, C(9,2)=36 $C(9,2)=36$ a square too, so the first two n's $n$'s are 2,9.
One can also see from the solutions of 2x^2-y^2=-+1 $2x^2-y^2=\pm 1$ pell's equation, by take x=1, y=0 $x=1, y=0$ for n=2=2*1 $n=2=2*1$ and take x=2 y=3 $x=2,$y=3$for n=9=3^2, Thanks,$n=9=3^2\$.