SOLUTION: "if (2^k)-1 is a prime number then, prove that (2^(k-1))(2^k-1)is a perfect number." (i know it is the definition but i want proof)

Question 476630: "if (2^k)-1 is a prime number then, prove that (2^(k-1))(2^k-1)is a perfect number."
(i know it is the definition but i want proof)
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
Theorem:
If is a number, then is a number and every even perfect number has this form.
Proof:
Suppose first that is a number, and set .
To show is we need only show . Since is multiplicative and , we know
.
This shows that is a number.
On the other hand, is any number and write as where is an and .
Again sigma is multiplicative so
.
Since is we also know that
.
Together these two criteria give
,
so, divides hence divides , say
.
Now substitute this back into the equation above and by to get .
Since and are both of we know that
,
so .
This means that is and its two are itself () and one ().
Thus is a and we have that the number has the prescribed form.

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