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For all perfect numbers $N$, $\sigma (N) = 2N$, where $\sigma$ is the divisor sigma function.

Let $s$ be a perfect number of the form $3^m 5^n 7^k$, where $m,n,k \geq 1$ are integers.

Then $\sigma (s)= \sigma (3^m 5^n 7^k)$

$ =\sigma (3^m) \sigma (5^n) \sigma (7^k)$ since $3, 5,$ and $7$ are coprime to each other.

$ =\left(\frac{3^{m+1}-1}{2}\right)\left(\frac{5^{m+1}-1}{4}\right)\left(\frac{7^{k+1}-1}{6}\right)$

$ =2(3^m 5^n 7^k)$ since $s$ is a perfect number.

$\implies 9 (3^m 5^n 7^k) = 3^{m+1} 5^{n+1}+3^{m+1} 7^{k+1} + 5^{n+1} 7^{k+1} - 3^{m+1}-5^{n+1} - 7^{k+1}-1$ after some algebra.

This is as far as I got using this method. Any and all help would be appreciated.

hardmath
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Anonymous
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3 Answers3

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In order to get a perfect number, we need the ratio $\dfrac{\sigma(n)}{n}=2$, which also means that we need the prime powers in the denominator to appear in the sigma values on top. And we need exactly and only one power of $2$ to appear in that sigma value, because the denominator is odd.

Unrolling this into prime powers,

$$\frac{\sigma(n)}{n}=\frac{\sigma(3^m)}{3^m} \frac{\sigma(5^n)}{5^n} \frac{\sigma(7^k)}{7^k}$$

For primes in general, $p$ does not divide $\sigma(p^j)$. So we would need to match the powers of each prime from the $\sigma$ values of the other primes.

For $\sigma(3^m)$, any odd value of $m$ makes $\sigma(3^m)$ divisible by $4$, so we need $m$ even - but in that case, $\sigma(3^m)$ is not divisible by $5$ - we actually need $m{+}1$ divisible by $4$ to get any power of $5$, which can't happen.

Similarly, for $\sigma(7^k)$, any odd value of $k$ makes $\sigma(7^k)$ divisible by $8$, so we need $k$ even - but again in that case, $\sigma(7^k)$ is not divisible by $5$ which once again only happens if $k{+}1$ is divisible by $4$.

So a number of that form cannot have $\dfrac{\sigma(n)}{n}=2$ and thus cannot be perfect.

Joffan
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Intuitively speaking, the left hand side is generally way smaller than the right hand side because of all the exponentials, so diving by those exponentials, $3^{m+1}, 5^{n+1}, 7^{k+1}$ is a good way to exploit that behavior.

Then you basically play around to find that no values for $m, n$, and $k$ work. (I checked, but it seems that writing it up would solve it for you.)

user357980
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  • E.g. if you wanted to prove that there is no $n$ such that $4\times 3^n5^n = 3^{n+1} + 5^{m+1} - 1$ if you divide you get: $4/15 = 3/5^{n+1} + 5/3^{m+1} - 1/3^{m+1}5^{n+1}$, which then gives you $4/15 < 3/5^{n+1} + 5/3^{m+1}$, which you can find which $m$'s and $n$'s give you a contradiction, then you can restrict to those $m$'s and $n$'s in the original equation to get a contradiction. – user357980 Sep 02 '17 at 22:45
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    Please review How do I write a good Answer? While no one should feel coerced into "writing it up [as that] would solve it for you", that is the purpose of the Answer box. Math.SE is trying to collect and curate content that provides reasonably concise and coherent Questions and Answers at all levels of mathematical study. – hardmath Sep 02 '17 at 23:53
  • I have read the policy before and was trying to give a suggestion that encouraged the OP to solve the problem on their own... unlike the majority of people who post more-or-less full solutions, even when the OP just asks for a hint. I was not complaining about writing a full solution, I was saying that if I did then they won't have to solve it themself. I just reviewed the link and it seems to me that I did what it suggested. I am not exactly sure what your criticism is. I tried to give a motivation for what I did in the first paragraph and gave an example of the method in the comment below. – user357980 Sep 03 '17 at 01:27
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Note that the Question stipulates the exponents considered are $m,n,k\ge 1$. As @lulu points out, the cases where one of the exponents is zero can (if desired) be ruled out by this previous Question.

The following is a simplification of the proof that an odd perfect number cannot be divisible by $105$ found here, as previously linked under the older Question to that effect.

$N= 3^m 5^n 7^k$ is a perfect number if and only $S(N)$, the sum of all divisors of $N$ (including itself and $1$), equals $2N$.

Since $N$ is odd, it must be that $S(N)=2N$ is not divisible by $4$. Now:

$$ \frac{S(N)}{N} = \left(1+\frac{1}{3}+\ldots+\frac{1}{3^m}\right) \left(1+\frac{1}{5}+\ldots+\frac{1}{5^n}\right) \left(1+\frac{1}{7}+\ldots+\frac{1}{7^k}\right) $$

Since $m=1$ would give $\left(1+\frac{1}{3}+\ldots+\frac{1}{3^m}\right)=\frac{4}{3}$ and $k=1$ would give $\left(1+\frac{1}{7}+\ldots+\frac{1}{7^k}\right)=\frac{8}{7}$, either would imply $S(N)$ is divisible by $4$, contradicting our observation above.

Knowing thus $m,k\ge 2$, we get a contradiction:

$$ \begin{align*} 2 = \frac{S(N)}{N} &\ge \left(1+\frac{1}{3}+\frac{1} {3^2} \right) \left(1+\frac{1}{5}\right) \left(1+\frac{1}{7}+\frac{1}{7^2}\right) \\ &= \frac{13}{9} \frac{6}{5} \frac{57}{49} = \frac{4446}{2205} \gt 2 \end{align*} $$

hardmath
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