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)

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> 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)      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Theorem:
If 2k-1 is a prime number, then 2k-1%282k-1%29 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%282k-1%29.
To show n is perfect we need only show sigma%28n%29+=+2n. Since sigma is multiplicative and sigma%28p%29+=+p%2B1+=+2k, we know
sigma%28n%29+=+sigma%282k-1%29.sigma%28p%29+=++%282k-1%292k+=+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%3E2.
Again sigma is multiplicative so
sigma%282k-1m%29+=+sigma%282k-1%29.sigma%28m%29+=+%282k-1%29.sigma%28m%29.
Since n is perfect we also know that
sigma%28n%29+=+2n+=+2km.
Together these two criteria give
2km+=+%282k-1%29.sigma%28m%29,
so, 2k-1 divides 2km hence 2k-1 divides m, say
m+=+%282k-1%29M.
Now substitute this back into the equation above and divide by 2k-1 to get 2kM+=+sigma%28m%29.
Since m and M are both divisors of m we know that
2kM+=+sigma%28m%29+%3E+m+%2B+M+=+2kM,
so sigma%28m%29+=+m+%2B+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.