Constructing Triangular Numbers From Square Numbers
\[2n^2-1, \; 2n^2+1\]
is square (\[{}=m^2\]
)then \[(nm)^2\]
is triangular, since\[(nm)^2=n^2m^2=n^2(2n^2-1)=\frac{2n^2}{2}(2n^2-1)=\frac{k}{2}(k-1)=\frac{j}{2}(j+1)\]
where first
\[k=2n^2\]
then \[k=j+1\]
.or
\[(nm)^2=n^2m^2=n^2(2n^2+1)=\frac{2n^2}{2}(2n^2+1)=\frac{k}{2}(k+1)\]
where first
\[k=2n^2\]
.
