$n$
is said to be triangular if $n =\sum_{i=0}^{k}{i}$
for some positive integer $k.$
Given $8n+1$
is a square number, show that $n$
is triangular.

Solve
$\sum_{i=0}^{k}{i}.$

The condition can be written as
$8n+1 = m^2,$
for some natural$m$
. 
Manipulate the above to a form that allows factorization.

If
$(m1)(m+1)$
is divisible by$8,$
what values can$m$
take? 
Would
$(m1)(m+1)$
be divisible by$8$
if$m$
was even? 
Use a substitution for
$m$
to express that$m$
is odd.

A positive integer
$n$
is said to be triangular if$n =\sum_{i=0}^{k}{i}=\frac{k(k+1)}{2}.$
Since$8n+1$
is a square number, there exists a positive integer$m$
such that$8n + 1 = m^2.$
Using the difference between two squares, we can factorize to get$8n = m^21 = (m1)(m+1),$
or that$n=\frac{(m1)(m+1)}{8},$
which means that$(m1)(m+1)$
must be divisible by 8. This is true if and only if$m$
is odd, since for every consecutive even numbers, one is a multiple of$4.$
Substituting$m = 2k+1$
gives us$n = \frac{2k(2k+2)}{8} = \frac{k(k+1)}{2} = \sum_{i=0}^{k}{i}.$