Triangular Numbers and Perfect Squares 
Prove that $n$ is a triangular number if and only if $8n+1$ is a perfect square.

I proved the easier part first (I think), that is, if $n$ is a triangular number then $8n+1$ is a perfect square.
I don't know where to start from for the other part, please help.
By the way, this was taken from David Burton's book, Elementary Number Theory and sadly it doesn't have all the  solutions...

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*If $n$ is a triangular number, then $n = \frac{k(k+1)}{2}$ and $8n+1 = 4k(k+1)+1 = (2k+1)^2$.

*If $8n+1$ is a perfect square, then $8n+1 = (2m+1)^2 \implies n = \frac{4m^2+4m+1-1}{8} = \frac{m(m+1)}{2}$ (because $8n+1$ is odd so it must be square of an odd number). 
