$n$ 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.
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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.
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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.
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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$n = \frac{2k(2k+2)}{8} = \frac{k(k+1)}{2} = \sum_{i=0}^{k}{i}.$