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Let $$\tau(n) = \sum_{d \mid n}{1}$$ be the divisor function, $$\omega(n) = \sum_{p \mid n}{1}$$ be the prime divisor function, $$\varphi(n) = \#\{1 \leqslant k \leqslant n : \gcd(k,n) = 1\}$$ be Euler's totient function, and $$\sigma(n) = \sum_{d \mid n}{d}$$ be the classical sum-of-divisors function.

I am reading this preprint by Gabdullin and Iudelevich, where they showed that:

  • The map $m \to m\tau(m)$ is not injective. (They give the example $18\tau(18) = 27\tau(27)$, and therefore, $18t\cdot\tau(18t) = 27t\cdot\tau(27t)$ whenever $\gcd(t,6)=1$.)

  • The map $m \to m\omega(m)$ is also not injective. (For any prime $q \geqslant 5$, we have $18q = 9q\omega(9q) = 6q\omega(6q)$.)

  • The map $m \to m\varphi(m)$ is an injection. (The details are in the first paragraph of Section 4 in page 8.)

So, now I wonder:

  • Is the map $m \to m\sigma(m)$ an injection?

I ran the following Pari-GP script in Sage Cell Server to search for integers $u \neq v$ such that $$u\sigma(u) = v\sigma(v)$$ in the range $$1 \leqslant u \leqslant 100, 1 \leqslant v \leqslant 100,$$ the computer search found:

12[2, 2; 3, 1]14[2, 1; 7, 1]
14[2, 1; 7, 1]12[2, 2; 3, 1]
48[2, 4; 3, 1]62[2, 1; 31, 1]
60[2, 2; 3, 1; 5, 1]70[2, 1; 5, 1; 7, 1]
62[2, 1; 31, 1]48[2, 4; 3, 1]
70[2, 1; 5, 1; 7, 1]60[2, 2; 3, 1; 5, 1]
  • This shows that the map $m \to m\sigma(m)$ is not injective, as we have $$336=12\sigma(12)=14\sigma(14),$$ $$5952=48\sigma(48)=62\sigma(62),$$ and $$10080=60\sigma(60)=70\sigma(70).$$

Note that all known examples $u, v$ below $100$ are even. Extending the search until $u, v \leqslant 1000$, and limiting it to values of $u$ and $v$ such that $uv$ is odd, then we obtain

315[3, 2; 5, 1; 7, 1]351[3, 3; 13, 1]
351[3, 3; 13, 1]315[3, 2; 5, 1; 7, 1]
  • This means that $$196560=315\sigma(315)=351\sigma(351).$$

The Pari-GP interpreter of Sage Cell Server crashes as soon as search limits of $u, v \leqslant {10}^4$ are specified.

Here are my:

QUESTIONS

(1) Are there more examples of positive integers $u \neq v$ such that $uv$ is odd and $$u\sigma(u)=v\sigma(v)?$$

(2) If the answer to Question (1) is YES, are there infinitely many such examples?

(3) Are there any examples of positive integers $u \neq v$ such that $uv$ is odd, $\gcd(u,v)=1$, and $$u\sigma(u)=v\sigma(v)?$$

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    $\begingroup$ FYI, A337874 does not have such an example. $\endgroup$
    – mathlove
    Commented Jan 29, 2022 at 14:05
  • $\begingroup$ Thank you for your time and attention, @mathlove! Will take your lead to investigate more into this problem (and the accompanying OEIS sequences). $\endgroup$ Commented Jan 29, 2022 at 14:13

2 Answers 2

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infinitely many odd pairs $$ u = 3^2 \cdot 5 \cdot 7 \cdot w$$ $$ v = 3^3 \cdot 13 \cdot w $$ for odd $w$ such that $$ \gcd(w, 2 \cdot 3 \cdot 5 \cdot 7 \cdot 13) = 1. $$

After running much higher, it now appears that these are all the successful pairs. Worth trying to prove, easier tasks would include ruling out $u=p$ prime, $u = p q $ semiprime.

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  • $\begingroup$ And for $p=11$. $\endgroup$
    – paw88789
    Commented Jan 29, 2022 at 17:55
  • $\begingroup$ @paw88789 yes, indeed $p$ can be anything relatively prime to $3 \cdot 5 \cdot 7 \cdot 13 = 1365$ $\endgroup$
    – Will Jagy
    Commented Jan 29, 2022 at 18:10
  • $\begingroup$ Thank you for your time and attention, @WillJagy! +1 $\endgroup$ Commented Jan 30, 2022 at 5:27
  • $\begingroup$ @WillJagy: Did you mean "$w=p$ prime, $w=pq$ semiprime" instead of "$u=p$ prime, $u=pq$ semiprime" in your last sentence? $\endgroup$ Commented Jan 30, 2022 at 5:37
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    $\begingroup$ @ArnieBebita-Dris no, I meant to start over, drop the 3, 13, 5, 7 factors, see if the easiest apparent restrictions can be proved. Long road to prove (if true) that my pairs are all the examples..... $\endgroup$
    – Will Jagy
    Commented Jan 30, 2022 at 14:32
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I can answer question 1 in the affirmative.

For example $3465\cdot \sigma(3465)=3861\cdot \sigma(3861)$

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    $\begingroup$ I found several more examples; all seem to have a common factor of $9$ (relevant to your question 3). $\endgroup$
    – paw88789
    Commented Jan 29, 2022 at 17:46
  • $\begingroup$ Thank you for your time and attention, @paw88789! +1 $\endgroup$ Commented Jan 30, 2022 at 5:27

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