A positive integer $n$ 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 $(m-1)(m+1)$ is divisible by $8,$ what values can $m$ take?

Would $(m-1)(m+1)$ 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 $8n = m^2-1 = (m-1)(m+1),$ or that $n=\frac{(m-1)(m+1)}{8},$ which means that $(m-1)(m+1)$ 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 $n = \frac{2k(2k+2)}{8} = \frac{k(k+1)}{2} = \sum_{i=0}^{k}{i}.$