The problem is to find all n for which n choose 2 is a perfect square. In other words, when m is an integer.

I'm not really at all sure how to approach this, any help is appreciated.

$\displaystyle {n \choose 2} = \frac{n!}{2! ( n -2)!} = \frac {n(n-1)}{2} = m^2 $

Also, here are the 10 smallest solutions

n = 2 m = 1

n = 9 m = 6

n = 50 m = 35

n = 289 m = 204

n = 1682 m = 1189

n = 9801 m = 6930

n = 57122 m = 40391

n = 332929 m = 235416

n = 1940450 m = 1372105

n = 11309769 m = 7997214