You are currently browsing the tag archive for the ‘egyptian fractions’ tag.

I have just uploaded to the arXiv my paper “On the number of solutions to “, submitted to the Journal of the Australian Mathematical Society.

For any positive integer , let denote the number of solutions to the Diophantine equation

where are positive integers (we allow repetitions, and do not require the to be increasing). The Erdös-Straus conjecture asserts that for all . By dividing through by any positive integer we see that , so it suffices to verify this conjecture for primes , i.e. to solve the Diophantine equation

for each prime . As the case is easily solved, we may of course restrict attention to odd primes.

This conjecture remains open, although there is a reasonable amount of evidence towards its truth. For instance, it was shown by Vaughan that for any large , the number of exceptions to the Erdös-Straus conjecture with is at most for some absolute constant . The Erdös-Straus conjecture is also verified in several congruence classes of primes; for instance, from the identity

we see that the conjecture holds when . Further identities of this type can be used to resolve the conjecture unless is a quadratic residue mod , which leaves only six residue classes in that modulus to check (namely, , and ). However, there is a significant obstruction to eliminating the quadratic residue classes, as we will discuss later.

By combining these reductions with extensive numerical calculations, the Erdös-Straus conjecture was verified for all by Swett.

One approach to solving Diophantine equations such as (1) is to use methods of analytic number theory, such as the circle method, to obtain asymptotics (or at least lower bounds) for the number of solutions (or some proxy for this number); if one obtains a lower bound which is nontrivial for every , one has solved the problem. (One can alternatively view such methods as a variant of the probabilistic method; in this interpretation, one chooses the unknowns according to some suitable probability distribution and then tries to show that the probability of solving the equation is positive.) Such techniques can be effective for instance for certain instances of the Waring problem. However, as a general rule, these methods only work when there are a *lot* of solutions, and specifically when the number of solutions grows at a polynomial rate with the parameter .

The first main result of my paper is that the number of solutions is essentially logarithmic in nature, thus providing a serious obstruction to solution by analytic methods, as there is almost no margin of error available. More precisely, I show that for almost all primes (i.e. in a subset of primes of relative density ), one has , or more precisely that

Readers familiar with analytic number theory will recognise the right-hand side from the divisor bound, which indeed plays a role in the proof.

Actually, there are more precise estimates which suggest (though do not quite prove) that the mean value of is comparable to . To state these results properly, I need some more notation. It is not difficult to show that if is an odd prime and obey (1), then at least one, but not all, of the must be a multiple of . Thus we can split , where is the number of solutions to (1) where exactly of the are divisible by . The sharpest estimates I can prove are then

Theorem 1For all sufficiently large , one hasand

Since the number of primes less than is comparable to by the prime number theorem, this shows that the mean value of in this range is between and , and the mean value of is between and , which gives the previous result as a corollary (thanks to a Markov inequality argument). A naive Poisson process heuristic then suggests that each prime has a “probability” of having a solution to (1), which by the Borel-Cantelli lemma heuristically suggests that there are only finitely many for which (1) fails. Of course, this is far from a rigorous proof (though the result of Vaughan mentioned earlier, which is based on the large sieve, can be viewed as a partial formalisation of the argument).

The first step in obtaining these results is an elementary description of in terms of some divisibility constraints. More precisely, we have

Lemma 2Let be an odd prime. Then is equal to three times the number of triples of positive integers, with coprime, dividing , and dividing . Similarly, is equal to three times the number of triples of positive integers, with coprime, dividing , and dividing .

One direction of this lemma is very easy: if divides and divides then the identity

and its cyclic permutations give three contributions to , and similarly if divides instead then the identity

and its cyclic permutations give three contributions to . Conversely, some elementary number theory can be used to show that all solutions contributing to either or are of these forms. (Similar criteria have been used in prior literature, for instance in the previously mentioned paper of Vaughan.)

This lemma, incidentally, provides a way to quickly generate a large number of congruences of primes for which the Erdös-Straus conjecture can be verified. Indeed, from the above lemma we see that if are coprime and is any odd factor of (and hence coprime to ), then the Erdös-Straus conjecture is true whenever or . For instance,

- Taking , we see that the conjecture holds whenever (leaving only those primes );
- Taking we see that the conjecture holds whenever (leaving only those primes );
- Taking , we see that the conjecture holds whenever ((leaving only those primes );
- Taking , we see that the conjecture holds whenever (leaving only those primes );
- Taking , we see that the conjecture holds whenever ; taking instead , we see that the conjecture holds whenever . (This leaves only the primes that are equal to one of the six quadratic residues , an observation first made by Mordell.)
- etc.

If we let (resp. ) be the number of triples with coprime, dividing , and congruent to (resp. ) mod , the above lemma tells us that for all odd primes , so it suffices to show that are non-zero for all . One might hope that there are enough congruence relations provided by the previous observation to obtain a covering system of congruences which would resolve the conjecture. Unfortunately, as was also observed by Mordell, an application of the quadratic reciprocity law shows that these congruence relations cannot eliminate quadratic residues, only quadratic non-residues. More precisely, one can show that if are as above (restricting to the case , which contains all the unsolved cases of the conjecture) and is the largest odd factor of , then the Jacobi symbols and are always equal to . One consequence of this is that whenever is an odd perfect square. This makes it quite hard to resolve the conjecture for odd primes which resemble a perfect square in that they lie in quadratic residue classes to small moduli (and, from Dirichlet’s theorem on primes in arithmetic progressions, one can find many primes of this type). The same argument also shows that for an odd prime , there are no solutions to

in which one or two of the are divisible by , with the other denominators being coprime to . (Of course, one can still get representations of by starting with a representation of and dividing by , but such representations are is not of the above form.) This shows that any attempt to establish the Erdös-Straus conjecture by manually constructing as a function of must involve a technique which breaks down if is replaced by (for instance, this rules out any approach based on using polynomial combinations of and dividing into cases based on residue classes of in small moduli). Part of the problem here is that we do not have good bounds that prevent a prime from “spoofing” a perfect square to all small moduli (say, to all moduli less than a small power of ); this problem is closely related (via quadratic reciprocity) to Vinogradov’s conjecture on the least quadratic nonresidue, discussed in this previous blog post.

It remains to control the sums for . The lower bounds are relatively routine to establish, arising from counting the contribution of those that are somewhat small (e.g. less than ), and use only standard asymptotics of arithmetic functions and the Bombieri-Vinogradov inequality (to handle the restriction of the summation to primes). To obtain (nearly) matching upper bounds, one needs to prevent from getting too large. In the case of , the fact that divides and divides soon leads one to the bounds , at which point one can count primes in residue classes modulo with reasonable efficiency via the Brun-Titchmarsh inequality. This inequality unfortunately introduces a factor of , which we were only able to handle by bounding it crudely by , leading to the double logarithmic loss in the sum.

For , these arguments do not give good bounds, because it is possible for to be much larger than , and one can no longer easily count solutions to . However, an elementary change of variables resolves this problem. It turns out that if one lets be the positive integers

then one can convert the triple to a triple obeying the constraints

Conversely, every such triple determines by the formulae

The point is that this transformation now places in a residue class modulo rather than , and with , this allows one to count the number of solutions reasonably efficiently. The main loss now comes from counting the number of divisors of ; in particular, one becomes interested in estimating sums such as

where are less than and is the number of divisors of . In principle, because behaves like on the average (as can be seen from the Dirichlet hyperbola method), we expect this sum to be of order , but I was unable to show this, instead using the far cruder divisor bound

to obtain an upper bound of . Any improvement upon this bound would lead to a corresponding improvement in the upper bound for . While improvement is possible in some ranges (particularly when is large) using various bounds on Kloosterman-type exponential sums, I was not able to adequately control these sums when was fairly small (e.g. polylogarithmic size in ), as one could no longer extract much of an averaging effect from the summation in that case. Part of the difficulty is that in that case one must somehow exploit the fact that is irreducible as a polynomial in for any fixed , otherwise there will be too many divisors.

## Recent Comments