Summation vs Square

$\displaystyle \sum_{k=1}^Nk=m^2$

or, $\frac {N(N+1)}{2}=m^2$

or, $N(N+1)=2m^2$

We have to find the least integer $N > 1$ such that $N(N+1)= 2m^2$ where m is a positive integer. We shall distinguish two cases:

$1$ . $N=2a$ . $a$ is a positive integer.Our equation takes on the form $a(2a+1)=m^2$

The factors on the left-hand side are pairwise relatively prime, hence they all must be squares. If $a=1$ ,then $2a+1$ is not a square.The next square after $1$ is $4$ . If $a=4$ , $2a+1=3^2$ .Then $N=2a=8$

$2$ . $N=2a+1$ . $a$ is a positive integer.Our equation takes on the form $(a+1)(2a+1)=m^2$

The factors on the left-hand side are pairwise relatively prime, hence they all must be squares.The least $a$ for which $2a+1$ is square is $a=4$ .But then $N=2a+1=9>8$ .

Thus $\boxed{8}$ is the least value of $N>1$