Prove that if $2^{p}-1$ is prime then $$n=2^{p-1}(2^p-1)$$ is a perfect number
here is what i did:
We need to prove the $\sigma(n)=n$
so $\sigma(n)=\sigma(2^{p-1})\sigma(2^p-1)$
since $2^{p}-1$ is a prime thus $\sigma(2^p-1)=2^p$
since $2$ is prime we have $\sigma(2^{p-1})=\frac{2^p-1}{2-1}=2^p-1$
so we have $\sigma(n)= 2^p(2^p-1)\neq n$
someone please help where did i go wrong?
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6We need to prove that $\sigma(n)=2n$. And you have. – André Nicolas Jun 05 '14 at 21:03
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oh right forgot about that – H_Hassan Jun 05 '14 at 21:05
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This is exercise 4.2.16 of Niven, Zuckerman, Montgomery's An Introduction to the Theory of Numbers 5th Ed. – 19021605 Oct 27 '25 at 11:56
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Let $p \geq 2$ and that $d=2^p-1$ is a prime.
Then,the divisors of $2^{p-1} d$ are these: $$1,2, \dots, 2^{p-1},d,2d, \dots , 2^{p-1} d$$
Therefore,
$$\sigma(n)=1+2+ \dots+ 2^{p-1}+d+2d+ \dots+ 2^{p-1} d=(1+2+ \dots+ 2^{p-1})(1+d) \\ =(2^p-1)2^p=2n$$
So,we conclude that $n$ is a perfect number.
evinda
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A perfect number is one whose sum of divisors is twice the number. For the given $n$, the sum of the factors is $\sum_{r=0}^{p-1}\: ^nC_r \cdot 2^r \cdot (2^p - 1) = 2^p \cdot (2^p -1) = 2n$. Thus, proved that n is a perfect number.
user3628041
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$\sum_{r=0}^{p-1}\binom{n}{r}2^r(2^p-1)=(2^p-1)(1+2)^{p-1}=(2^p-1)3^{p-1}$. – 19021605 Oct 27 '25 at 11:59