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Confused as to how to show the number 6's uniqueness. This theorem/problem comes from the projects section of "Reading, writing, and proving" from Springer.

Definition 1. The sum of divisors is the function $$\sigma (n) = \sum_{d\,|\,n} d,$$ where $d$ runs over the positive divisors of $n$ including 1 but not $n$ itself.

Definition 2. The product of divisors is the function $$p(n) =\prod_{d\,|\,n} d,$$ where $d$ runs over the positive divisors of $n$ including 1 but not $n$ itself.

This is my progress, ends very quickly:

So the logical form of the problem is $\exists!x \left( \sigma(x) = x \wedge p(x) = x \right)$, which can be reexpressed as either $\exists x((\sigma(x) = x \wedge p(x) = x) \wedge \forall y (\sigma(y) = y \wedge p(y) = y)\rightarrow y=x)$ or $ \exists (\sigma(x) = x \wedge p(x) = x) \wedge \forall y \forall z ((\sigma(y) = y \wedge p(y) = y) \wedge (\sigma(z) = z \wedge p(z) = z) \rightarrow y=z)$. We use existential instantiation and choose x to be 6. So, choosing the first method — this choice seemed simpler to me — $(\sigma(6)= 6 \wedge p(6) = 6) \wedge \forall y ( \sigma(y)=y\wedge p(n) = n) \rightarrow y=6$. How do we go on about proving 6's uniqueness; how do we get that y=6?

Theorem. There is only one positive integer that is both the product and sum of all its proper positive divisors, and that number is $6$.

Proof.

Existence: Suppose $n =6$. Then $\sigma(6) = 1 + 2 + 3 =6 $ and $p(6)= 1 \cdot 2 \cdot 3 = 6$, so 6 is both the product and sum of all its proper positive divisors.

Uniqueness: [I have no idea.] $\square$

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    $\begingroup$ First off: You are approaching this problem way too formally. This only introduces unneccessary baggage you have to carry with you. If you’re gonna travel the lands of number theory, travel lightly. $\endgroup$ – k.stm Jul 16 at 10:42
  • $\begingroup$ The sum of the positive proper divisors of $n$ is not $\sigma(n)$, but $\sigma(n) - n$. Therefore, $n$ is equal to the sum of its positive proper divisors if and only if $$\sigma(n)=2n.$$ Such positive integers $n$ are called perfect numbers. $\endgroup$ – Arnie Bebita-Dris Jul 22 at 4:19
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Let $n$ be a positive integer that satisfies the requirement. It can be readily checked that $n>1$ and $n$ is not a prime power.

If $n=p^k$ for some prime natural number $p$ and for some positive integer $k$, then we have $$1+p+p^2+\ldots+p^{k-1}=p^k=1\cdot p\cdot p^2\cdot \ldots\cdot p^{k-1}\,.$$ Thus, $p$ divides $1+p+p^2+\ldots+p^{k-1}$. Do you see a problem here?

Therefore, $n$ has at least two distinct prime factors. Let $p$ and $q$ denote two prime distinct natural numbers that divide $n$. Obviously, $pq\mid n$, whence $$n\geq pq\,.$$

Then, $\dfrac{n}{p}$, $\dfrac{n}{q}$, and $\dfrac{n}{pq}$ are proper divisors of $n$. Consequently, as $n$ is the product of its (positive) proper divisors, we get $$n\geq \left(\dfrac{n}{p}\right)\cdot\left(\dfrac{n}{q}\right)\cdot\left(\dfrac{n}{pq}\right)=\frac{n^3}{p^2q^2}\,.$$ Therefore, $n^2\leq p^2q^2$, or $n\leq pq$. However, $n\geq pq$. We then conclude that $n=pq$.

Thus, $1$, $p$, and $q$ are the only positive proper divisors of $n$. Ergo, from the requirement, $$1\cdot p\cdot q=n=1+p+q\,.$$ Therefore, $pq=p+q+1$, or $$(p-1)(q-1)=2\,.$$ You can finish this, I suppose.


Related Questions.

(a) If $n$ is a positive integer such that $n$ equals the product of all positive proper divisors of $n$, then show that $n=p^3$ for some prime natural number $p$, or $n=pq$ for some distinct prime natural numbers $p$ and $q$.

(b) If $n$ is a positive integer such that the product of all positive proper divisors of $n$ equals the sum of all positive proper divisors of $n$ (without requiring that the product or the sum is equal to $n$ itself), then prove that $n=6$.

(c) If $n$ is a positive integer such that the product of all positive divisors of $n$ equals the sum of all positive divisors of $n$, then prove that $n=1$.

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We seek $n \in \mathbb N$ satisfying

$$ \prod_{d \mid n} d = n^2 \:\:\text{and}\:\: \sigma(n) = \sum_{d \mid n} d = 2n. \quad \ldots \quad (\star) $$

If $d(n)$ denotes the number of (positive) divisors of $n$, then

$$ (n^2)^2 = \left( \prod_{m \mid n} m \right) \left( \prod_{m \mid n} \tfrac{n}{m} \right) = \prod_{m \mid n} n = n^{d(n)}. $$

Therefore, $d(n)=4$, and $n=p^3$, $p$ prime, or $n=pq$, $p,q$ distinct primes.

Now $\sigma(p^3)=1+p+p^2+p^3 \ne 2p^3$, since $p \nmid \sigma(p^3)$. Finally,

$$ 2pq = \sigma(pq) = (p+1)(q+1) = pq+(p+q)+1 $$

is equivalent to $(p-1)(q-1)=2$. This implies $p=2$ and $q=3$, so $n=6$ is the only integer satisfying the condition in $(\star)$. $\blacksquare$

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