Let $\mu\left(n\right)$ be the Möbius function. Let $\phi\left(n\right)$ be Euler's totient function. Let $\sigma\left(n\right)$ be the sum of divisors and $\tau\left(n\right)$ be the number of divisors functions. I am curious to know whether or not the system:





has at most one solution.

Motivation: I remember a number theory assignment I had where we were given particular values for each of these functions and asked to recover the original number. I can't for the life of me remember how (or if) I managed to solve this problem. I tried to work out a general proof, but couldn't. I also wrote a loop in maple to check for counterexamples, but haven't found any yet. I feel like this is something I should know, but probably have forgotten the relevant facts to approaching this problem.

  • $\begingroup$ For small values of a, b, c, d it's just casework on the prime factorization of n. That's probably how you did it. $\endgroup$ Commented Nov 22, 2010 at 18:52
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    $\begingroup$ Now that we've found that the answer is "no", the next question is can we add more number-theoretic functions so that the answer becomes "yes". I believe that each pair of Alon's counterexamples have different values of big-omega(n), the number of prime factors of n counted with multiplicity. $\endgroup$ Commented Nov 22, 2010 at 18:56
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    $\begingroup$ My guess, based on no evidence whatsoever, is that it'll be hard to come up with a reasonable list of functions to make the function injective. Of course the function "n" would be really helpful here... but honest number theoretic functions admit lots of collisions. $\endgroup$
    – Alon Amit
    Commented Nov 22, 2010 at 19:05
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    $\begingroup$ Well, "product of prime divisors (with repetition)" will do it on its own, as well :-) I guess the super-vague claim is that the standard functions mu, phi, tau etc. won't do. $\endgroup$
    – Alon Amit
    Commented Nov 22, 2010 at 19:20
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    $\begingroup$ The value of $n \phi(n)$ uniquely determines $n$, so one function is all you need. $\endgroup$ Commented Nov 22, 2010 at 23:27

2 Answers 2


The answer is No. The smallest counterexamples I could find are {1836, 1824), {5236, 4960}, {5742, 5112}, {6764, 6368}, {9180, 9120} and {9724, 9184}. I think those are all the pairs in which both numbers are less than 10,000.

For example, both $n=1836$ and $n=1824$ satisfy $\mu(n)=0$, $\varphi(n)=576$, $\sigma(n)=5040$ and $\tau(n)=24$.

EDIT: here's the code of the program I used in GAP.

vec := function(n)
 return [MoebiusMu(n), Phi(n), Sigma(n), Tau(n)];

dic:=NewDictionary([1,2,3,4], true);

for i in [2..10000] do
 if (LookupDictionary(dic, v) <> fail) then Print(i," <=> ", LookupDictionary(dic, v), "\n"); fi;
 AddDictionary(dic, v, i);
  • $\begingroup$ Out of curiosity, how did you write your search algorithm? I haven't had to program in a long time and I was trying to write one in maple. I set up a for-loop but couldn't figure out how to compare the output data without inspecting the list myself. $\endgroup$
    – WWright
    Commented Nov 22, 2010 at 18:35
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    $\begingroup$ I made it super-easy for myself: I used GAP, so all the number-theoretic functions were available. In order to test for collisions, I used a hash table; GAP has built-in functions for that, too (I actually used GAP's "dictionary" concept). The complete code is like 8 lines. $\endgroup$
    – Alon Amit
    Commented Nov 22, 2010 at 18:42
  • $\begingroup$ Thanks for sharing the code! I might try modifying this myself to see what number theoretic properties the multiple solutions have in common. $\endgroup$
    – WWright
    Commented Nov 22, 2010 at 19:05

First, congratulations on asking a number theory question which is quite unlike any I have seen before. It's certainly more fun to read about questions like this rather than computing $a^b$ modulo $n$.

Regarding the assignment you got: one can certainly find particular values of $a$, $b$, $c$ and $d$ such that there is exactly one $n$ solving the equations. For instance, if $p$ is a prime number then taking $a = -1$, $b = p-1$, $c = p+1$, $d = 2$ has the unique solution $n = p$: $\tau(n) = 2$ forces $n$ to be prime and thus $\varphi(n) = n-1$.

Although I am a number theorist, I don't have an expert opinion on the general question. (I'm not really "the right kind" of number theorist to answer this question. You might ask Carl Pomerance, for instance.) I suppose the more reasonable guess is that there are values of $a,b,c,d$ which yield more than one solution.

Probably the first thing to do is what you're doing: search for counterexamples by computer. When you get tired of doing that, let us know how far you looked (and, of course, whether you've found any). The second thing to do is to come up with a heuristic on how often one might expect multiple solutions to be found...

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    $\begingroup$ I would vote up this answer multiple times if I could. Incidentally, my opinion (as a probabilist who is amused by number-theoretic heuristics) is that examples should get rarer as the numbers involved get larger, but I won't hazard a guess off the top of my head as to whether the number of counterexamples is finite or infinite. $\endgroup$ Commented Nov 22, 2010 at 19:20
  • $\begingroup$ Now the original question has been answered, perhaps Pete, WWright or Alon could post the natural follow up question concerning the frequency of multiple solutions as no doubt many of you are already wondering about this; I know I am. (I don't want to post it, as I wasn't involved in this question before this comment.) I expect the follow up question may not yield so readily. $\endgroup$ Commented Nov 22, 2010 at 20:24
  • $\begingroup$ @Derek Jennings - I was hesitant to post the follow up until I'd thought about the problem on my own a bit more. I also thought it might be inappropriate to ask multiple questions on the same topic in a short period of time. If you have a precise way of writing out what it means to define the frequency of multiple solutions, I'd encourage you to post the question. I find my own background a bit to modest to formulate the question very precisely, but am certainly curious about the answer :) $\endgroup$
    – WWright
    Commented Nov 22, 2010 at 20:58
  • $\begingroup$ @WWright As you're ok with that, and I'm certainly curious about the answer too (although sadly I've no time to work on it at the moment), I've just posted the follow up question. $\endgroup$ Commented Nov 22, 2010 at 22:08

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