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(This question is related to the latest comment underneath this MSE question.)

Let $$\sigma(x)=\sum_{d \mid x}{d}$$ be the sum of the divisors of $x$.

Denote the abundancy index of $x$ by $$I(x)=\frac{\sigma(x)}{x}.$$

Here is my question:

If $I(x)=\dfrac{p+2}{p}$, where $p$ is an odd prime, does it follow that $x$ is an odd square?

From the equation $$p\sigma(x)=(p+2)x$$ I can only prove that $x$ is an odd square when $x$ is odd. But what about if $x$ is even?

That is, if $I(x) = \dfrac{p+2}{p}$, where $p$ is an odd prime, must it necessarily be the case that $x$ is odd?

Ѕᴀᴀᴅ
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Jose Arnaldo Bebita Dris
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2 Answers2

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First, note that for any coprime $a, b \in \mathbb{N}_+$, there is$$ I(ab) = I(a) I(b). $$

Suppose there is an even number $n = 2^k \cdot l$, where $k \geqslant 1$ and $l$ odd, such that$$ I(n) = \frac{p + 2}{p}. $$

Case 1: $k \geqslant 2$. Then$$ \frac{p + 2}{p} = I(2^k \cdot l) = I(2^k) I(l) \geqslant I(2^k) = \frac{2^{k + 1} - 1}{2^k} = 2 - \frac{1}{2^k} \geqslant \frac{7}{4}, $$ which implies $4(p + 2) \geqslant 7p$, contradictory to $p \geqslant 3$.

Case 2: $k = 1$ and $p \geqslant 5$. Then, analogously,$$ \frac{p + 2}{p} \geqslant 2 - \frac{1}{2^k} = \frac{3}{2}, $$ which implies $2(p + 2) \geqslant 3p$, contradictory to $p \geqslant 5$.

Case 3: $k = 1$ and $p = 3$. Then$$ \frac{5}{3} = I(2l) = \frac{3}{2} I(l) \Longrightarrow I(l) = \frac{10}{9}. $$

If $l$ has an odd prime factor $q < 10$, suppose $q^m \mathrel{\|} l$, then$$ \frac{10}{9} = I(l) = I(q^m) I\left( \frac{l}{q^m} \right) \geqslant I(q^m) = \frac{q^{m + 1} - 1}{q^m (q - 1)}\\ = 1 + \frac{1}{q - 1}\frac{q^m - 1}{q^m} \geqslant 1 + \frac{1}{q - 1}\frac{q - 1}{q} = 1 + \frac{1}{q} > \frac{10}{9}, $$ a contradiction. Now suppose the prime factorization of $l$ is$$ l = \prod_{i = 1}^s p_i^{a_i}, $$ then $p_i \geqslant 11$. Because$$ \frac{10}{9} = I(l) = \prod_{i = 1}^s \frac{1 + p_i + \cdots + p_i^{a_i}}{p_i^{a_i}}, $$ then$$ 9 \prod_{i = 1}^s \sum_{j = 0}^{a_i} p_i^j = 10 \prod_{i = 1}^s p_i^{a_i} \Longrightarrow \left. 9 \,\middle|\, \prod_{i = 1}^s p_i^{a_i} \right., $$ contradictory to $p_i \geqslant 11$.

Therefore, there does not exist an even positive integer $n$ such that$$ I(n) = \frac{p + 2}{p}. $$

Ѕᴀᴀᴅ
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  • Thanks @AlexFrancisco, I honestly did not see this coming. Is this your own proof? Otherwise, care to share the online reference from where this proof is lifted? – Jose Arnaldo Bebita Dris Mar 05 '18 at 09:38
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    @JoseArnaldoBebitaDris This is my original proof. – Ѕᴀᴀᴅ Mar 05 '18 at 10:06
  • Dude, you should have that published. I do not remember seeing such a proof in the literature. – Jose Arnaldo Bebita Dris Mar 05 '18 at 10:25
  • Hi again, @AlexFrancisco. I seem to have spotted an error in your proof just now. In the part where you assumed that $l$ has a prime factor $q < 10$, which implies that $1/q > 1/10$, then $1 + (1/q) > 11/10$, which is not greater than $10/9$. – Jose Arnaldo Bebita Dris Mar 06 '18 at 04:38
  • But I think that your proof is leading to a hint that $q \leq 9$ (since $q$ must be prime), so that $$10/9 \geq I(q^m) \geq 1 + (1/q) \geq 10/9,$$ which implies that $I(q^m) = 10/9$. This contradicts the fact that $10/9$ is an abundancy outlaw (since $9 < 10 < \sigma(9) = 1 + 3 + 9 = 13$ and $9$ is relatively prime to $10$). See this paper by Holdener and Stanton for more. – Jose Arnaldo Bebita Dris Mar 06 '18 at 04:44
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    @JoseArnaldoBebitaDris If $q$ is a prime and $q < 10$, then $q < 9$ also holds. – Ѕᴀᴀᴅ Mar 06 '18 at 05:03
  • Hi @AlexFrancisco. Yes, you are right! I missed that! =) – Jose Arnaldo Bebita Dris Mar 06 '18 at 05:09
  • Hi again, @AlexFrancisco. We would like to include your proof here as additional material content for our paper, submitted to NNTDM. In this regard, may we invite you as a co-author? If you do not wish to become a co-author, may we have your permission to include your proof in our paper? You may reach me at josearnaldobdris AT gmail DOT com. Thank you! =) – Jose Arnaldo Bebita Dris Jun 28 '18 at 11:46
  • Hi @AlexFrancisco. I sent you an e-mail. Please check. Thank you! =) – Jose Arnaldo Bebita Dris Oct 09 '18 at 09:18
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p must divide x and so we can suppose that 2p is a factor of x. Then I(2p) is no greater than (p+2)/p which is clearly false.

  • +1 for the simple proof! =) – Jose Arnaldo Bebita Dris Aug 21 '19 at 13:07
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    Thanks Jose. You may also be interested to note that a similar reasoning applies to your problem 1472823 where your initial gcds being 1 would imply that 3 divides y and then the index would have to be bigger than 4/3. –  Aug 21 '19 at 15:34
  • I invite you to write your last comment out as a complete answer to that other question, @S.Dolan, so that I will be able to accept it. =) – Jose Arnaldo Bebita Dris Aug 22 '19 at 03:35
  • Just double-checking your argument, @S.Dolan: Assume that $x$ is even. Then we have: $$p \mid x \text{ and } x \text{ even } \implies 2p \mid x \implies \frac{3}{2}\cdot\frac{p+1}{p} = I(2)I(p) = I(2p) \leq I(x) = \frac{p+2}{p}$$ from which it follows that $$3(p+1) \leq 2(p+2) \implies 3p + 3 \leq 2p + 4 \implies p \leq 1,$$ which contradicts the fact that $p$ is an odd prime. – Jose Arnaldo Bebita Dris Sep 15 '19 at 18:35
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    Yes, that's exactly what I considered. –  Sep 15 '19 at 18:58