Then \(T=\frac12n(n+1)\) for some whole number \(n\)
Let \(T\) be a triangular number
Therefore, if \(T\) is triangular, \(8T+1\) is square
We wish to prove that if \(T\) is a triangular number then \(8T+1\) is a square number.
Factorising the right hand side, \(8T+1=(2n+1)^2\)
Therefore \(8T+1 = 8(\frac12n(n+1))+1\)