Let \(N\) be a *Perfect Number* (a positive integer that is equal to the sum of its proper positive divisors), like \(6\), or \(28\). It seems that all known perfect numbers are the sum of a series of consecutive positive integers starting with \(1\), that is, it seems that any perfect number is a **triangular number**. For example:

\(6=1+2+3\),

\(28=1+2+3+4+5+6+7\).

Prove that any **even** perfect number is triangular.

Note: It is not known whether there are any odd perfect numbers.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestAny even perfect number can be expressed as \(2^{n-1}(2^n-1)\) for some natural number \(n\).Now it's quite easily seen that \(2^{n-1}(2^n-1)=\dfrac{2^n(2^n-1)}{2}\) which fits the expression \(\dfrac{k(k+1)}{2}\) (this is the expression for all triangular numbers) when \(k=2^n-1\)

Log in to reply