## Perfect Triangles

#### Problem

The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of 28 are 1, 2, 4, 7, 14, and 28, so the sum of proper divisors is 1 + 2 + 4 + 7 + 14 = 28.

The first eight perfect numbers are 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128.

It can be shown that P is an even perfect number iff it is of the form $2^{n-1}(2^n - 1)$ where $2^n - 1$ is prime.

Prove that all even perfect numbers are triangle numbers.

#### Solution

By definition the $m$th triangle number, $T_m = 1 + 2 + 3 + ... + m = \dfrac{m(m+1)}{2}$.

Let $m = 2^n - 1$.

\begin{align}\therefore T_m &= \dfrac{(2^n - 1)(2^n - 1 + 1)}{2}\\&= \dfrac{(2^n - 1)2^n}{2}\\&= 2^{n-1}(2^n - 1)\end{align}

Hence all even perfect numbers are triangle numbers.

Furthermore, $T_m$ is a perfect number if $m$ is a prime of the form $2^n - 1$. For example, $2^2 - 1 = 3$ and $T_3 = 6$; $2^3 - 1 = 7$ and $T_7 = 28$; $2^5 - 1 = 31$ and $T_31 = 496$; and so on.

Problem ID: 329 (13 Jul 2007)     Difficulty: 3 Star

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