Each friend of 10 has at least 10 nonidentical prime factors

For each positive integer n, if the sum of the factors of n is divided by n, then the result is called the abundancy index of n. If the abundancy index of some positive integer m equals the abundancy index of n but m is not equal to n, then m and n are called friends. A positive integer with no friends is called solitary. The smallest positive integer that is not known to have a friend and is not known to be solitary is 10. It is not known if the number 6 has odd friends, that is, if odd perfect numbers exist. In a 2007 article, Nielsen proved that the number of nonidentical prime factors in any odd perfect number is at least 9. A 2015 article by Nielsen, which was more complicated and used a computer program that took months to complete, increased the lower bound from 9 to 10. This work applies methods from Nielsen's 2007 article to show that each friend of 10 has at least 10 nonidentical prime factors. This is a formal write-up of results presented at the Southern Africa Mathematical Sciences Association Conference 2023 at the University of Pretoria.

If m and n are positive integers, then m is called a friend of n if m = n and σ −1 (m) = σ −1 (n); a positive integer n is called solitary if no friend of n exists.This terminology appeared in a problem in the American Mathematical Monthly, where it was conjectured that the set of solitary numbers has density zero [1].The smallest positive integer that is not known to have a friend and is not known to be solitary is 10; it seems, although it has not been proved, that for every prime p that is at least 5, the number 2p is solitary (see [9]).
It is not known if the number 6 has odd friends, that is, if odd perfect numbers exist.In a 2007 article [7], Nielsen proved that the number of nonidentical prime factors in any odd perfect number is at least 9.A 2015 article by Nielsen [8], which was more complicated and used a computer program that took months to complete, increased the lower bound from 9 to 10.
This work applies methods from Nielsen's 2007 article [7] to show that each friend of 10 has at least 10 nonidentical prime factors.This is a formal write-up of results presented in a talk at the Southern Africa Mathematical Sciences Association Conference 2023, held at the University of Pretoria, Pretoria, South Africa, in November 2023.

Theory
The following definitions are used for positive integers m, positive integers n, and primes p: the notation m | n means "m divides n", the notation m | n means "m does not divide n", the number v p (n) is the largest nonnegative integer v such that p v | n, the integer v p (m/n) is v p (m) − v p (n), and the number ω(n) is the number of nonidentical prime factors of n.
Some well-known facts about σ −1 are recalled.
Lemma 1.Let p and q be primes, and let a, b, m, and n be positive integers.The following foundational result underpins the later conclusions of this paper.
Proposition 2. Let n be a friend of 10.
(a) The number n is a square, n is divisible by 5, and n is coprime to 6.(b) The number n has at least five nonidentical prime factors.
Let n = k j=1 p aj j where the numbers p j are different primes, the numbers a j are positive integers, and , so every number a j is even, so n is a square.
For a contradiction, suppose 3 | n.It follows that p 1 = 3, p 2 = 5, and 2 , and m are pairwise relatively prime, so , which is impossible.Therefore, n is not a multiple of 3.
If n has at most four nonidentical prime factors, then which is impossible.Therefore, n has at least five nonidentical prime factors.
Remarks.For odd perfect numbers n (that is, odd positive integers n such that σ −1 (n) = 2), similar arguments yield the following conclusions: (a) n is of the socalled Eulerian form p a m 2 where m and a are positive integers, p is prime, p does not divide m, and p ≡ a ≡ 1 (mod 4); and (b) n has at least three nonidentical prime factors (because If p is a given prime such that p ≥ 7, then for friends n of 2p (that is, positive integers n such that σ −1 (n) = (3/2)(p+1)/p), similar arguments yield the following more trivial conclusions: (a) n is odd and divisible by p, and (b) n has at least two nonidentical prime factors (because 3/2 < (3/2)(p + 1)/p < (3/2)(5/4)).Imposing the additional condition p ≡ 1 (mod 4) yields the result that n is a square, but does not improve the conclusion about ω(n).
The reason why the arguments above easily restrict ω(n) for friends of 10 but not for friends of 2p where p = 5 is: in the case of friends of 10, but apparently not in other cases, it can easily be proved that 3 does not divide n.Thus, the friends-of-10 problem seems suited to computer search to an extent that other similar problems are not.
The following result will be applied in computer calculations; to obtain it, a wellknown argument (see [5, section 2] and [7, section 6]) is adapted to find bounds on an unknown prime factor of n, where σ −1 (n) is known exactly or known to be in some given interval.
(In applications, the primes p 1 through p k1 are known, the exponents a 1 through a ℓ1 are known, but the other primes p j and the other exponents a j are not known; however, the lower bounds b j are known.)Suppose that the positive real numbers t min and t max satisfy .
It follows that Proof.(a) Since k 1 < k, the hypotheses imply If a k1+1 ≥ 2, then since k 1 < k, the hypotheses imply .
The result with this rational function of M is used instead of the result p k1+1 > 1/(2 √ M − 2) in order to maintain exact arithmetic when performing computer calculations.In practice, the previous result 1/(M − 1) ≤ p k1+1 is improved by the additional factor of 8/((2 − M ) 2 + 7), since that factor is larger than 1 in the case where 1 , from which the last required inequality follows since m > 1.
(d) Use part (c) and then part (b) to obtain For positive integers m and n such that gcd(m, n) = 1, let o n (m) be the order of m modulo n, that is, the smallest positive integer c such that m c ≡ 1 mod n.Each of the following two propositions is similar to a proposition in the 2007 paper of Nielsen [7, section 3]; to make the current article as self-contained as possible, brief proofs are given here.Proposition 4. Let a be a positive integer, let p be an odd prime, and let x be an integer such that x > 1 and p | x. (a) Proof.If m is an integer such that m > 1 and p | m − 1, then v p (m p − 1) = v p (m − 1) + 1, because p is odd and by the binomial theorem, If m is an integer such that m > 1 and p | m − 1, and the positive integer r is a nonmultiple of p, then v p (m r − 1) = v p (m − 1), because by the binomial theorem, The results of the last two sentences imply that if m is an integer such that m > 1 and p | m − 1, and r is a positive integer, then v p (m r − 1) = v p (m − 1) + v p (r).Each part of the result now follows from the fact that (For part (a), take m = x; for part (b), take m = x op(x) and note that o p (x), being a factor of p − 1, is a nonmultiple of p.) Proposition 5. Let p be an odd prime, and let a be an even positive integer.It follows that for each factor d of a + 1 such that d > 1, the number σ(p a ) has a prime factor q d such that o q d (p) = d.
Proof.Note that (x a+1 − 1)/(x − 1) = d:d>1 and d|a+1 Φ d (x), where Φ d (x) is the dth cyclotomic polynomial.By a result of Bang [2], for each integer d > 1 and each prime p, the number Φ d (p) has a prime factor q such that the order of p modulo q is d, except if (d, p) = (1, 2), (d, p) = (6, 2), or (d = 2 and p + 1 is a power of 2); none of the three exceptions occurs if both d and p are odd.
To the author's knowledge, the following two corollaries are new.
Let f be the number of prime factors q of n such that q = r and o r (q) | v q (n) + 1. (This includes the prime factors q of n such that r | q − 1.)By Proposition 4, f ≥ 1 since v r (σ(n)) ≥ 1.By the generalised pigeonhole principle, there is some prime factor q of n other than r such that o r (q) | v q (n) + 1 and , so the numbers r, r 2 , r 3 , . . ., r b are factors of v q (n)+ 1, so by Proposition 5, σ(q vq (n) ) has at least b different prime factors q r , q r 2 , . . ., q r b such that for each i ∈ {1, . . ., b}, the following results hold: o q r i (q) = r i , so Corollary 7.For every friend n of 10, Proof.Apply Corollary 6 with r = 5 and c = 0; note that This section ends with a proposition that will be used to speed up a computer program in two situations where, apparently, the run time would otherwise be prohibitively long.First, recall the following well-known lemma (which is equivalent to [7,Lemma 11]).Lemma 8. Let p be an odd prime, let a be a positive integer, and let y be an integer coprime to p.The equation z p−1 ≡ 1 mod p a has exactly one solution z such that z ≡ y mod p.That solution satisfies z ≡ y p a−1 mod p a .Proof.For the existence and uniqueness of z, apply Hensel's lemma to X p−1 − 1 ≡ p−1 y=1 (X − y) mod p.Now (zy −1 ) p a−1 ≡ 1 mod p a (applying Proposition 4(a) if a > 1), so y p a−1 ≡ z p a−1 ≡ z mod p a .
For a given odd prime p and a given positive integer a, it follows that to search for integers x coprime to p such that p a−1 < x ≤ p a and v p (x p−1 − 1) ≥ a, it is enough to check the numbers y p a−1 mod p a for y ∈ {2, . . ., p − 1}.(Note that 1 p a−1 ≡ 1 mod p a .)Carrying out that check by computer for the case p = 31, a ≤ 15 and the case p = 19531, a ≤ 7 yields the following result, which is a slight improvement of [7,Lemma 12] applicable to fewer cases than that lemma.Proposition 9.If some integer x is coprime to 31 and satisfies 1 < x ≤ 31 14 , then If some integer x is coprime to 19531 and satisfies 1 < x ≤ 19531 6 , then

Computer program
The factor-chain-search scheme used by Nielsen [7] is adapted.The core idea is: for positive integers n and primes p, if v p (n) = a > 0, then for each prime factor q of σ(p a ), if v q (σ −1 (n)) ≤ 0 then v q (n) ≥ v q (σ(n)) > 0, so q | n.In this way, a prime factor p of n, together with a known exponent a, can generate other prime factors q of n under mild conditions.For example, for a friend n of 10, if A SageMath computer program was implemented; the program code and output files are included as supplemental files attached to this paper. 1A detailed description of the program follows.
The program finds candidate partial prime factorisations for all positive integers n such that • ω(n) = k, and • Every prime p that satisfies v p (σ −1 (n)) > 0 is in S ignore , where t min and t max are user-specified rational numbers greater than 1, where k is a user-specified positive integer, and where S ignore is a user-specified finite list of primes which this paper calls ignored primes.The user specifies a bound B, which is a cutoff value above which powers of primes are considered to be "large".
The program performs a depth-first search of a tree of cases.At the start of each branch of the tree: • There is a known on sequence S on , which consists of finitely many known distinct prime factors of n, which are called the on primes; • For each prime q in S on , it is known that v q (n) = a q or it is known that v q (n) ≥ b q , where a q or b q respectively is a known positive integer; • There is a known off sequence S off , which consists of finitely many known distinct prime factors of n, each of which is not in S on ; the primes in S off are called the off primes; • For each prime q in S off , it is known that v q (n) ≥ b q , where b q is a known positive integer; and • A number P is known such that for every prime factor q of n, if q is neither in S on nor in S off , then q > P .The values of S on , S off , the corresponding exponents a q and b q , and P at the start of the program -that is, at the root of the tree -are specified by the user.The on and off primes (respectively, their exact exponents or minimum exponents) are some prime factors (respectively, their exact exponents or nonstrict lower bounds for their exponents) in a potential number n.The (exact or minimum) exponents of the on primes have been finalised; the exponents of the off primes have not yet been finalised.Each time the program moves from one level of the tree to the next level, the number of on primes increases by exactly 1, and the number of off primes may change.The off primes are thought of as being a by-product of the on primes.
The user may choose to specify, or not to specify, a special prime r such that v r (σ −1 (n)) = 0; that prime, if specified, is available for the program to use in * If every prime factor of σ(p a ) is greater than P or in S ignore or in S on or in S off , then start a new branch of the tree with the data obtained from the old branch's data as follows: move p from S off to S on , let p have the exact exponent a p = a, and for every prime factor q of σ(p a ) that is neither in S ignore nor in S on , do the following: if q is in S off , then increase the minimum exponent b q of q by v q (σ(p a )); otherwise, append q to the end of S off and let it have minimum exponent b q = v q (σ(p a )).* Increase a by 2.
-Start a new branch of the tree with the data obtained from the old branch's data by moving p from S off to S on and letting p have the minimum exponent b p = a.• Otherwise, if no "do not proceed further" instruction has been encountered in this iteration of the program, then do the following.
-If m ≤ 1, then indicate that there is no upper bound for the next prime and do not proceed further along this branch of the tree.-For each prime p in the interval I = (max{P, B low }, B high ) (going through those primes p in ascending order), if p is not in S on then do the following.(If there are no primes in I, then do not proceed further along this branch of the tree.) ≤ L, and a r or b r is strictly larger than (k − 1) 2 + q∈Son∪S off f r (q) + ⌈log r B high ⌉ + ⌈log r g(p)⌉ + 2δ, then break out of the p for loop (Corollary 6 is violated for all primes p yet to be checked in this loop: g is strictly decreasing and the loop goes through the primes p in ascending order).* Let a = 2. * While p a ≤ B, do the following.
• Find the prime factorisation of σ(p a ).
• If every prime factor of σ(p a ) is greater than P or in S ignore or in S on , then start a new branch of the tree with the data obtained from the old branch's data as follows: append p to the end of S on , let p have the exact exponent a p = a, let the new value of P be p, and for every prime factor q of σ(p a ) that is neither in S ignore nor in S on , do the following: if q is in S off , then increase the minimum exponent b q of q by v q (σ(p a )); otherwise, append q to the end of S off and let it have minimum exponent b q = v q (σ(p a )).• Increase a by 2. * Start a new branch of the tree with the data obtained from the old branch's data by appending p to the end of S on , letting p have the minimum exponent b p = a, and letting the new value of P be p.
The SageMath program was run repeatedly, according to the specifications in Table 1.Each line of the table that includes a CPU time refers to one run or to k S on a q and b q for q in S on B (r, log r L, δ) (CPU time)/s 5 Empty 10 • If a line of the table imposes a condition on the value of the variable a, where some on prime p is listed as having exponent a p = a, then one run was done for each even-positive-integer value of a satisfying the given conditions, and the CPU time in that line refers to all of those runs combined.At the start of each of those runs, S off was taken to be the increasing sequence consisting of the different prime factors q of σ(p a ) such that q > 5, with the following exceptions. of σ(19531 72 ) = (19531 73 − 1)/19530 for the case a = 72.In all cases (including the exceptions above, where trial division or Fac-torDB was used), at the start of the run, for each off prime q, the exponent b q was taken to be v q (σ(p a )).
• If a line of the table does not mention a variable a, then the line refers to a single run and, at the start of that run, S off was taken to be empty.• If a line of the table has a blank space in the (r, log r L, δ) column, then the run(s) used no special prime.The numbers t min and t max were taken to be 9/5.The sequence S ignore was taken to be the singleton sequence (3).The number P was taken to be 4 for k ≤ 7 and 5 for k ≥ 8.
The values of the bound B were chosen by experimentation; that process of trial and error initially used a Magma version of the program on the online Magma calculator [3], which apparently limits computation time to 60 seconds per run.Where " * " appears in Table 1, the values of B were as follows.

Results
Each run of the program terminated without errors and found no candidate partial prime factorisations.The runs with 5 ≤ k ≤ 7 rule out friends n of 10 such that 5 ≤ ω(n) ≤ 7. By Corollary 6 and Corollary 7, the runs with 8 ≤ k ≤ 9 rule out friends n of 10 such that 8 ≤ ω(n) ≤ 9. Therefore, the main theorem of this paper is proved: Theorem 10.Each friend of 10 has at least 10 nonidentical prime factors.

Remarks concerning run time
Apparently, Corollary 7 is very important for the proof, because the Corollary is needed to ensure that the runs for k = 9 finish in a reasonable amount of time.The runs for 5 ≤ k ≤ 7 did not use Corollary 7, and each of those runs took less than a second.However, although a run that tackled k = 8 in the same way as 5 ≤ k ≤ 7 (using S on = ∅, using B = 10 29 , using P = 4, using no special prime, and without using Corollary 7) was successful, it took 28500 seconds to complete.The author believes that it is futile to try doing k = 9 like this.Indeed, in two cases with k = 9, Corollary 7 was not enough to reduce the run time to a manageable duration, and the results regarding special primes also needed to be used.(Many thanks to an anonymous referee for inquiring about the importance of Corollary 7.) The Cunningham Project tables [4] confirm that for even positive integers a ≤ 124, the number σ(5 a ) has at least two different prime factors, except if a ∈ {2, 6, 10, 12, 46}, in which case σ(5 a ) is prime.As Table 1 illustrates, each of the five cases (ω(n), v 5 (n)) ∈ {(9, 2), (9,6), (9, 10), (9, 12), (9, 46)} takes several hours or uses the time-saving special prime r, whereas all other runs combined take less than a minute without using the special prime r.The five cases v 5 (n) ∈ {2, 6, 10, 12, 46} would most likely be the bottlenecks for any potential attempt, using this approach, to investigate friends n of 10 such that 10 ≤ ω(n) ≤ 12. (In order to use the 33-digit number σ(5 46 ) as the special prime r in the case v 5 (n) = 46, a version of Proposition 9 would be needed.To prove such a statement, a direct check of all possible remainders modulo σ(5 46 ) would not be feasible.) As an anonymous referee kindly pointed out, the analogous run times for the three runs with (ω(n), v 5 (n)) ∈ {(9, 10), (9,12), (9,46)} indicate that many calculations could be common to the three runs.This is indeed true: the large prime σ(5 v5(n) ) contributes little to σ(n) in all three runs, so significant sections of the trees of cases are the same in all three runs apart from the value of that prime.(For example, consider the first few levels within the branch of the tree where σ(5 v5(n) ) is moved from S off to S on at the first step.)However, the three runs are not completely identical (so simply exchanging the three runs for one run representing v 5 (n) ≥ 10 is not possible), and they were sufficiently fast for the purposes of this paper.In general, as the referee also kindly pointed out, the program takes advantage of being able to use a specific value of σ(5 v5(n) ) to generate tree branches.
In the runs with (k, a 5 ) = (9, 2) where a 31 = a ≤ 94, it appears that most of the time was used to factorise σ(31 a ), which is why trial division was used to avoid having to factorise σ(19531 a ) completely in the runs with (k, a 5 ) = (9,6) where a 19531 = a ∈ [30, 86] and a / ∈ {58, 72}.The numbers σ(19531 58 ) and σ(19531 72 ) are composite but have no prime factors less than 2 31 (the largest allowed finite upper limit in the factor_trial_division routine in SageMath), which is why FactorDB was used to find the smallest prime factor of each of these two numbers.

1
SageMath version 10.0 (release date May 20, 2023, using Python 3.11.4)was run on Conda in Mambaforge using the Windows Subsystem for Linux on Windows 11, on a laptop with a 12th Gen Intel(R) Core(TM) i5-1235U CPU ("base speed" 1,30 GHz with many cores; in the run with ω(n) = 9, v 5 (n) = 2, and v 31 (n) ≤ 94, Windows Task Manager showed a speed of about 3,5 GHz).* Find the prime factorisation of σ(p a ).

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In the runs where k = 9, S on =(5, 19531), a 5 = 6, a 19531 = a, 30 ≤ a ≤ 86, and a / ∈ {58, 72}: instead of laboriously factoring σ(19531 a ) completely, trial division was used to express σ(19531 a ) as a product m j=1 p cj j where the primes p j satisfy p 1 < p 2 < • • • < p m < 2 31 , or as a product ( m j=1 p cj j )c where the primes p j and the positive integer c satisfy p 1 < p 2 < • • • < p m < 2 31 < c; in both cases, the numbers c j are positive integers.The sequence S off was taken to be the sequence obtained from (p 1 , . . ., p m ) by removing all terms p j such that p j ≤ 5.-In the two runs where k = 9, S on = (5, 19531), a 5 = 6, and a 19531 = a ∈ {58, 72}: S off was taken to be the singleton sequence (q), where the query 19531^n-1 in the FactorDB database[6] provided the 15-digit smallest prime factor q = 316636168836007 of σ(19531 58 ) = (19531 59 − 1)/19530 for the case a = 58, as well as the 26-digit smallest prime factor q = If p = 5, then gcd(σ(p a ), 30) = 1 since a is even, so S off includes every prime factor of σ(p a ).

Table 1 .
Input-related values and CPU times in computer searches for friends n of 10. multiple runs of the SageMath program.The second and third columns of the table indicate the value of S on and the corresponding exponents at the start of each run.