# 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. – k.stm Jul 16 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. – Arnie Bebita-Dris Jul 22 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 $$(\star)$$. $$\blacksquare$$