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

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.] $$\square$$

• 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. Commented Jul 16, 2020 at 10:42
• 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. Commented Jul 22, 2020 at 4:19

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 $$n=1$$.

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 $$(\star)$$. $$\blacksquare$$