Science:Math Exam Resources/Courses/MATH437/December 2011/Question 06

MATH437 December 2011

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Question 06
If $\displaystyle n\in \mathbb {N}$ is an even perfect number, then there exists a $\displaystyle k\in \mathbb {N}$ such that $\displaystyle 2^{k+1}-1$ is a prime number and $\displaystyle n=2^{k}(2^{k+1}-1)$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Let $\displaystyle n=2^{k}m$. First, recall $\displaystyle \sigma (n)=\sum _{d\mid n}d$ is a multiplicative function. The condition that n is an even perfect number translates to $\displaystyle 2n=\sigma (n)$.

Hint 2
Show that $\displaystyle 2^{k+1}-1\mid m$ so that $\displaystyle m=(2^{k+1}-1)M$ for some integer M.

Hint 3
A final piece of advice, notice that both m and M are divisors of m. Hence $\displaystyle \sigma (m)\geq m+M$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $\displaystyle n=2^{k}m$ with k at least 1. Further, let $\displaystyle \sigma (n)=\sum _{d\mid n}d$ and recall that this function is multiplicative (the sum is over positive divisors of n). Since we are given that n is an even perfect number, we know that $\displaystyle 2n=\sigma (n)$. Next, we use the multiplicativity of $\displaystyle \sigma$ and see that $\displaystyle 2^{k+1}m=2n=\sigma (n)=\sigma (2^{k}m)=\sigma (2^{k})\sigma (m)=(1+2+2^{2}+...+2^{k})\sigma (m)=(2^{k+1}-1)\sigma (m)$. Since $\displaystyle \gcd(2^{k+1},2^{k+1}-1)=1$, we see that $\displaystyle (2^{k+1}-1)\mid m$. Let M be an integer such that $\displaystyle m=(2^{k+1}-1)M$. Substituting this into the above yields $\displaystyle 2^{k+1}(2^{k+1}-1)M=2^{k+1}m=(2^{k+1}-1)\sigma (m)$. Cancelling on both sides yields $\displaystyle 2^{k+1}M=\sigma (m)$. Using the fact that both m and M are divisors of m (and hence are terms inside the expansion of $\displaystyle \sigma (m)$, we have that $\displaystyle 2^{k+1}M=\sigma (m)\geq m+M=(2^{k+1}-1)M+M=2^{k+1}M$. Thus, the greater than sign is in fact an equality. This means that $\displaystyle \sigma (m)$ has only two factors, namely m and M. Since $\displaystyle m>M$, we must have that $\displaystyle M=1$ and so $\displaystyle m=(2^{k+1}-1)M=2^{k+1}-1$. As m has only two prime factors, we also know that $\displaystyle m=2^{k+1}-1$ is prime. This completes the proof.

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