Question 476630
Theorem:
    If {{{2k-1}}} is a {{{prime}}} number, then {{{2k-1(2k-1)}}} is a {{{perfect}}} number and every even perfect number has this form.

Proof:  

Suppose first that  {{{p = 2k-1}}} is a {{{prime}}} number, and set {{{n = 2k-1(2k-1)}}}.  

To show {{{n}}} is {{{perfect}}} we need only show {{{sigma(n) = 2n}}}.  Since {{{sigma}}} is multiplicative and {{{sigma(p) = p+1 = 2k}}}, we know

    {{{sigma(n) = sigma(2k-1).sigma(p) =  (2k-1)2k = 2n}}}.

This shows that {{{n}}} is a {{{perfect}}} number.

    On the other hand, {{{suppose}}}{{{ n}}} is any {{{even}}}{{{ perfect}}} number and write {{{n}}} as {{{2k-1m}}} where {{{m}}} is an {{{odd}}}{{{ integer}}} and {{{k>2}}}.  

Again sigma is multiplicative so

    {{{sigma(2k-1m) = sigma(2k-1).sigma(m) = (2k-1).sigma(m)}}}.

Since {{{n}}} is {{{perfect}}} we also know that

    {{{sigma(n) = 2n = 2km}}}.

Together these two criteria give

    {{{2km = (2k-1).sigma(m)}}},

so, {{{2k-1}}} divides {{{2km}}} hence {{{2k-1}}} divides {{{m}}}, say 

{{{m = (2k-1)M}}}.  

Now substitute this back into the equation above and {{{divide}}} by {{{2k-1}}} to get {{{2kM = sigma(m)}}}.  

Since {{{m}}} and {{{M}}} are both {{{divisors}}} of {{{m}}} we know that

    {{{2kM = sigma(m) > m + M = 2kM}}},

so {{{sigma(m) = m + M}}}.  

This means that {{{m}}} is {{{prime}}} and its {{{only}}} two {{{divisors}}} are itself ({{{m}}}) and one ({{{M}}}).  

Thus {{{m = 2k-1}}} is a {{{prime}}} and we have {{{prove}}} that the number {{{n}}} has the prescribed form.