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I was working through some basic Number Theory Problems when I came across :

Given an integer $m$ $≥ 2$ such that $(2^{m} -1)$ is a prime, and $n$ $=$ $(2^{m-1})$$(2^{m} -1)$, then show that $\sigma(n)$ $=$ $2n$ where $\sigma(n)$ is the sum of the divisors of an integer $x$

I am not able to make progress ,can someone help me out ; even a hint will suffice

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  • $\begingroup$ For a hint, note that $\sigma$ is multiplicative, so $\sigma(n)=\sigma(2^{m-1})\sigma(2^m-1)$, Both of these are easy to compute. $\endgroup$ – André Nicolas Apr 6 '15 at 5:51
  • $\begingroup$ Taking $m=2$ gives $n=6$, and $6$ is indeed a perfect number, the smallest one. The OR is (was) wrong, $\endgroup$ – André Nicolas Apr 6 '15 at 5:54
  • $\begingroup$ Oh , @AndréNicolas , sorry - I was calculating something else :P $\endgroup$ – pranav Apr 6 '15 at 5:57
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    $\begingroup$ That is actually the Euler-Euclid Theorem for even perfect numbers $\geq 6$. We know that for all even perfect numbers $n$, $\sigma(n)=2n$. Hence follows your answer. You can see the proofs there. $\endgroup$ – Prasun Biswas Apr 6 '15 at 6:05
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    $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Apr 6 '15 at 6:08
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For $2^{m-1}$, the only numbers that divide it are of form $2^k$ over $0 \leq k \leq m-1$, so $\sigma(2^{m-1}) = \sum_{k=0}^{m-1} 2^k = 2^m-1$ via geometric series.

For prime $2^m-1$ (also known as a Mersenne prime), the only divisors it has are $1$ and itself, so $\sigma(2^m-1) = 2^m$ in this case.

Since $\sigma$ is multiplicative, we have $\sigma(ab) = \sigma(a)\sigma(b)$ where $\gcd(a,b)=1$. Therefore $\sigma((2^{m-1})(2^m-1)) = \sigma(2^{m-1})\sigma(2^m-1) = (2^m-1)(2^m) = 2n$.

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