This is a quick proof based on an observation inspired by “Mathematical Lens” in the May 2016 *Mathematics Teacher* (“Fence Posts and Rails” by Roger Turton). A triangular number is the sum of all integers from 1 to n. The general formula for T(n), the nth triangular number, is \[T(n) = \frac{(n)(n + 1)}{2}\]

**Challenge:** Prove that all triangular numbers are either multiples of three, or one more than a multiple of three. (That is, there are no triangular numbers that are one less than a multiple of three.)

We have three possible cases: \(n\) is a multiple of 3, \(n + 1\) is a multiple of 3, or \(n + 2\) is a multiple of 3. Let \(k\) represent some other integer. If the first case, \(n = 3k\), so \((n)(n + 1) = 3k(3k + 1)\), which is obviously a multiple of 3. In the second case, \(n + 1 = 3k\), so \((n)(n + 1) = 3k(3k – 1)\), which is obviously a multiple of 3.

This leaves the third case, in which neither \(n\) nor \(n + 1\) is a multiple of 3. In this case, since \(n + 2 = 3k\), then \(n = 3k – 2\) and \(n + 1 = 3k – 1\). This means that \(2T(3k-2) = (3k – 1)(3k – 2) = 9k^2 – 9k + 2 = 9k(k – 1) + 2\). Hence, \(T(3k-2) = 9\frac{k(k – 1)}{2} + 1\). The first term here is a multiple of 3. Thus: \[T(3k) = 3\frac{k(3k + 1)}{2} \\ T(3k-1) = 3\frac{k(3k – 1)}{2} \\ T(3k-2) = 3\frac{3k(k-1)}{2} + 1\]

Q.E.D.