(Note: This question has now been cross-posted to MO.)

Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors of $z$ by $s(z):=\sigma(z)-z$.

If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the form $n = p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Starting from the fundamental equality $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}$$ one can derive $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \gcd(m^2, \sigma(m^2))$$ so that we ultimately have $$\frac{D(m^2)}{s(p^k)} = \frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k} = \gcd(m^2, \sigma(m^2))$$ and $$\frac{s(m^2)}{D(p^k)/2} = \frac{\sigma(m^2) - m^2}{p^k - \frac{\sigma(p^k)}{2}} = \gcd(m^2, \sigma(m^2)),$$ whereby we obtain $$\frac{D(p^k)D(m^2)}{s(p^k)s(m^2)} = 2.$$ Note that we also have (Equation A) $$\frac{2D(m^2)s(m^2)}{D(p^k)s(p^k)} = \bigg(\gcd(m^2, \sigma(m^2))\bigg)^2.$$ Lastly, notice that we can easily get $$\sigma(p^k) \equiv k + 1 \equiv 2 \pmod 4$$ so that it remains to consider the possible equivalence classes for $\sigma(m^2)$ modulo $4$. Since $\sigma(m^2)$ is odd, we only need to consider two.

Here is my question:

Which equivalence class of $\sigma(m^2)$ modulo $4$ makes Equation A untenable?

  • $\begingroup$ Note that $\sigma(m^2)/p^k = \frac{m^2}{\sigma(p^k)/2}$ is a square if and only if $\sigma(p^k)/2$ is also a square. $\endgroup$ May 13, 2023 at 22:42

3 Answers 3


(The following answer is copied verbatim from the corresponding MO link.)

The following assertion appears in Theorem 3.3 (page 7, equations (5) to (6)) of Odd multiperfect numbers by Shi-Chao Chen and Hao Luo:

Let $n=\pi^{\alpha} M^2$ be an odd $2$-perfect number, with $\pi$ prime, $\gcd(\pi,M)=1$, and $\pi \equiv \alpha \equiv \pmod 4$. Then $$\sigma(M^2) \equiv 1 \pmod 4 \iff \pi \equiv \alpha \pmod 8,$$ $$\sigma(M^2) \equiv 3 \pmod 4 \iff \pi \equiv \alpha + 4 \pmod 8.$$


I cannot answer your question, but I proved two claims which might be useful to answer your question.

This answer proves the following claims :

Claim 1 : In $\text{mod}\ 4$, $$\sigma(m^2)\equiv 1\iff \frac{m^2}{\gcd(m^2, \sigma(m^2))}\equiv \frac{\sigma(m^2)}{\gcd(m^2, \sigma(m^2))}\iff \text{$\dfrac{2p^k-\sigma(p^k)}{4}\ $ is even}$$

Claim 2 : $$\sigma(m^2)\equiv \frac{(t+1)t}{8}\bigg(4+(t^2+1)(v-1)\bigg)\pmod 8$$ where $p\equiv t\pmod{16}$ and $k\equiv v\pmod{16}$.

(Though Claim 2 is "weaker" than what you've already known, it may be useful since all the cases are included in one expression.)

Claim 1 : In $\text{mod}\ 4$, $$\sigma(m^2)\equiv 1\iff \frac{m^2}{\gcd(m^2, \sigma(m^2))}\equiv \frac{\sigma(m^2)}{\gcd(m^2, \sigma(m^2))}\iff \text{$\dfrac{2p^k-\sigma(p^k)}{4}\ $ is even}$$

Proof :

In this proof, we consider in $\text{mod}\ 4$.

Let $$G= \gcd(m^2, \sigma(m^2)),\quad m^2=Ga,\quad\sigma(m^2)=Gb$$ where $a,b$ are odd such that $\gcd(a,b)=1$.

Then, we see that the equation A is equivalent to $$2D(m^2)s(m^2) = G^2D(p^k)s(p^k)$$ i.e. $$2(2Ga-Gb)(Gb-Ga)=G^2\bigg(2p^k-\sigma(p^k)\bigg)\bigg(\sigma(p^k)-p^k\bigg)$$ i.e. $$2(2a-b)(b-a)=\bigg(\underbrace{2p^k-\sigma(p^k)}_{\equiv\ 0}\bigg)\bigg(\underbrace{\sigma(p^k)-p^k}_{\equiv\ 1}\bigg)$$ i.e $$(2a-b)(b-a)=2c(4d+1)\tag1$$ where $c,d$ are integers such that $$2p^k-\sigma(p^k)=4c,\qquad \sigma(p^k)-p^k=4d+1$$

So, we have, from $(1)$, $$3ab-2a^2-b^2\equiv 2c\tag2$$ Since $a^2\equiv b^2\equiv 1$, we finally have, from $(2)$, $$ab\equiv 2c+1\tag3$$

Now, from $(3)$, we get the followings :

  • If $G\equiv a\equiv 1$ and $b\equiv 1$, then $\sigma(m^2)\equiv 1$, and $c$ is even.

  • If $G\equiv a\equiv 1$ and $b\equiv 3$, then $\sigma(m^2)\equiv 3$, and $c$ is odd.

  • If $G\equiv a\equiv 3$ and $b\equiv 1$, then $\sigma(m^2)\equiv 3$, and $c$ is odd.

  • If $G\equiv a\equiv 3$ and $b\equiv 3$, then $\sigma(m^2)\equiv 1$, and $c$ is even.

Therefore, we can say that

$$\sigma(m^2)\equiv 1\iff a\equiv b\iff \text{$c$ is even.}\quad\blacksquare$$

Claim 2 : $$\sigma(m^2)\equiv \frac{(t+1)t}{8}\bigg(4+(t^2+1)(v-1)\bigg)\pmod 8$$ where $p\equiv t\pmod{16}$ and $k\equiv v\pmod{16}$.

Proof :

We have

$$\sigma(m^2)\equiv p^km^2\bigg(\frac{\sigma(p^k)}{2}\bigg)^{-1}\equiv p^k\bigg(\frac{\sigma(p^k)}{2}\bigg)^{-1}\equiv p^k\bigg(\frac{\sigma(p^k)}{2}\bigg)\pmod 8$$

Let $p=16s+t\ (t>1), k=16u+v$ where $t^4\equiv 1\pmod{16}$.

Then, there is an integer $z$ such that $$\begin{align}\frac{\sigma(p^k)}{2}&\equiv \frac 12(1+t+\cdots +t^k)\pmod{8} \\\\&\equiv \frac{t^{k+1}-1}{2(t-1)}\pmod{8} \\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2(t^4)^{\frac{k-1}{4}}\bigg)\pmod{8} \\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2(t^4-1+1)^{\frac{k-1}{4}}\bigg)\pmod{8} \\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2\sum_{j=0}^{\frac{k-1}{4}}\binom{\frac{k-1}{4}}{j}(t^4-1)^j\bigg)\pmod{8} \\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2\bigg(1+\frac{k-1}{4}(t^4-1)+(t^4-1)^2z\bigg)\bigg)\pmod{8} \\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2+\frac{k-1}{4}(t^4-1)t^2\bigg)\pmod{8} \\\\&\equiv \frac{t+1}{2}+\frac{k-1}{2}t^2\frac{(t^2+1)(t+1)}{4}\pmod{8} \\\\&\equiv \frac{t+1}{2}+\frac{v-1}{2}t^2\frac{(t^2+1)(t+1)}{4}\pmod{8}\end{align}$$

Also, since $t^2\equiv 1\pmod 8$, $$p^k\equiv (16s+t)^{16u+v}\equiv t^v(t^4)^{4u}\equiv t^v\equiv t\pmod 8$$

Therefore, we finally get $$\sigma(m^2)\equiv \frac{(t+1)t}{8}\bigg(4+(t^2+1)(v-1)\bigg)\pmod 8$$ This holds even for $t=1$. $\ \blacksquare$

Examples :

If $p\equiv k\equiv 1\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.

If $p\equiv 5,k\equiv 13\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.

If $p\equiv k\equiv 9\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.

If $p\equiv 13,k\equiv 5\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.

Therefore, we can say that if $p+k\equiv 2\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$. (This is what you've already written in your answer.)


Further to this recent corrigendum in NNTDM, which corrects an oversight in Some modular considerations regarding odd perfect numbers – Part II, we realized that we do in fact have the following biconditionals:

In this article, we consider the various possibilities for $p$ and $k$ modulo $16$, and show conditions under which the respective congruence classes for $\sigma(m^2)$ (modulo $8$) are attained, if $p^k m^2$ is an odd perfect number with special prime $p$. We can prove that:

  • $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $p + k \equiv 2 \pmod {16}$
  • $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $p - k \equiv {12} \pmod {16}$
  • $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $p + k \equiv {10} \pmod {16}$
  • $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $p - k \equiv 4 \pmod {16}$.

If $k=1$, then these logical equivalences take the form

  • $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $p \equiv 1 \pmod {16}$
  • $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $p \equiv 13 \pmod {16}$
  • $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $p \equiv 9 \pmod {16}$
  • $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $p \equiv 5 \pmod {16}$.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .