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Would you please help me prove or disprove the conjecture highlighted below. I conceived it after solving a textbook problem to find distinct $n_1, n_2$ such that $n_1 \sigma(n_1) = n_2 \sigma(n_2)$ where $\sigma$ is the sum-of-divisors function. Out of curiosity, I used a table of the first one hundred thousand values of $\sigma$ and found 3408 such pairs: 12, 14; 48, 62; . . .; 94860, 95472. Out of more curiosity, I found 96 triples such that $n_1 \sigma(n_1) = n_2 \sigma(n_2) = n_3 \sigma(n_3)$: 336, 372, 434; . . .; 85680, 94860, 95472. Continuing, I found two quadruples such that $n_1 \sigma(n_1) = n_2 \sigma(n_2) = n_3 \sigma(n_3) = n_4 \sigma(n_4)$: 41664, 42672, 47244, 55118 and 42000, 46500, 51200, 54250.

A natural conjecture, then, is

For any positive integer $r$, there exist distinct positive integers $n_1, n_2, \dots, n_r$ such that $n_1 \sigma(n_1) = n_2 \sigma(n_2) = \cdots = n_r \sigma(n_r)$.

If the conjecture is true, then my search failed for $r \ge$ 5 only because my list of $\sigma$ values was not large enough.

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    $\begingroup$ True for r=5: {624960, 640080, 696384, 708660, 713232} is one of 3 solutions. Computed all the way through n = 10^7 (10 million, not just 1 million) and didn't find r=6. $\endgroup$
    – user2469
    Sep 7, 2017 at 19:45
  • $\begingroup$ This is true for amicable couples because n1 is equal to sum of divisors of n2 and n2 is equal to sum of divisors of n1. The 5 numbers found for r=5 must actually be amicable. It seems 6 amicable numbers can not exist. $\endgroup$
    – sirous
    Sep 24, 2017 at 14:24
  • $\begingroup$ @sirous Amicable numbers use the sum of proper divisors whereas $\sigma$ is the sum of all divisors. As a result, amicable numbers do not apply to this question. Note also that no more than two numbers can be amicable to each other because it is not possible for the sum of the proper divisors of a number to be equal to more than one number. $\endgroup$
    – user0
    Sep 25, 2017 at 16:53
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    $\begingroup$ Extended search to n=10^8 (100 million), still no result for r=6 $\endgroup$
    – user2469
    Oct 20, 2018 at 21:38

1 Answer 1

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The conjecture is true for all values up to at least $r = 51$. If, as is widely conjectured, there are infinitely many Mersenne primes, then the conjecture is true for all values of $r$.

First, a simple observation. If $n_1 \sigma(n_1) = n_2 \sigma(n_2)$ and $k$ is any integer coprime to both $n_1$ and $n_2$, then $n_1 k \sigma(n_1 k) = n_2 k \sigma(n_2 k)$. Think of $n = n_1 k$ as having a "replaceable part": if $n_1$ is replaced by $n_2$ yielding $n' = n_2 k$, this leaves $n' \sigma(n')$ the same as $n \sigma(n)$.

Mersenne primes provide a way to produce many examples of such replaceable parts. If $M(p) = 2^p - 1$ and $M(q) = 2^q - 1$ are two Mersenne primes, then let $n_1 = 2^{q-1} M(p)$ and $n_2 = 2^{p-1} M(q)$. Computing: $$ \begin{align} n_1 \sigma(n_1) &= 2^{q-1} M(p) \sigma(2^{q-1} M(p))\\ &= 2^{q-1} M(p)\, (2^q - 1) 2^p\\ &= 2^{p+q-1} M(p) M(q), \end{align} $$ and $n_2 \sigma(n_2)$ produces the same result.

Finally, one can construct a number with many replaceable parts as follows: Let $M(p_1),\ldots,M(p_r)$ be any set of $r$ distinct Mersenne primes, and let $n_1 = 2^{p_1 - 1} M(p_2)\cdots M(p_r)$ (note that $M(p_1)$ is omitted). For $i = 2,\ldots,r$, produce $n_i$ by replacing $2^{p_1 - 1} M(p_i)$ with $2^{p_i - 1} M(p_1)$. (That's hard on the eyes. A subscript of $1$ and a subscript of $i$ are interchanged.) This yields $r$ different values $n_i$ for which $n_i \sigma(n_i)$ has the same value.

(As it happens, the smallest examples of pairs, triples and quadruples listed in the question are produced in this fashion using the first two, three or four Mersenne primes.)

As of December $2018$, there are $51$ known Mersenne primes, so that proves the claim.

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