I’ve just uploaded to the arXiv my paper “On product representations of squares“. This short paper answers (in the negative) a (somewhat obscure) question of Erdös. Namely, for any , let be the size of the largest subset of with the property that no distinct elements of multiply to a square. In a paper by Erdös, Sárközy, and Sós, the following asymptotics were shown for fixed :
- .
- .
- .
- for .
- for .
- for .
In the end, the argument turned out to be relatively simple; no advanced results from additive combinatorics, graph theory, or analytic number theory were required. I found it convenient to proceed via the probabilistic method (although the more combinatorial technique of double counting would also suffice here). The main idea is to generate a tuple of distinct random natural numbers in which multiply to a square, and which are reasonably uniformly distributed throughout , in that each individual number is attained by one of the random variables with a probability of . If one can find such a distribution, then if the density of is sufficienly close to , it will happen with positive probability that each of the will lie in , giving the claim.
When , this strategy cannot work, as it contradicts the arguments of Erdös, Särközy, and Sós. The reason can be explained as follows. The most natural way to generate a triple of random natural numbers in which multiply to a square is to set
for some random natural numbers . But if one wants all these numbers to have magnitude , one sees on taking logarithms that one would need which by elementary linear algebra forces so in particular each of the would have a factor comparable to . However, it follows from known results on the “multiplication table problem” (how many distinct integers are there in the multiplication table?) that most numbers up to do not have a factor comparable to . (Quick proof: by the Hardy–Ramanujan law, a typical number of size or of size has factors, hence typically a number of size will not factor into two factors of size .) So the above strategy cannot work for .However, the situation changes for larger . For instance, for , we can try the same strategy with the ansatz
Whereas before there were three (approximate) equations constraining three unknowns, now we would have four equations and six unknowns, and so we no longer have strong constraints on any of the . So in principle we now have a chance to find a suitable random choice of the . The most significant remaining obstacle is the Hardy–Ramanujan law: since the typically have prime factors, it is natural in this case to choose each to have prime factors. As it turns out, if one does this (basically by requiring each prime to divide with an independent probability of about , for some small , and then also adding in one large prime to bring the magnitude of the to be comparable to ), the calculations all work out, and one obtains the claimed result.
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20 May, 2024 at 7:30 pm
Anonymous
Bravo!
20 May, 2024 at 7:53 pm
domotorp
See also https://arxiv.org/abs/2405.12088 which just appeared on arXiv.
20 May, 2024 at 8:08 pm
Terence Tao
Wow, that is quite a coincidence in timing! It seems the papers, while very adjacent in topic, do not actually overlap, but I will certainly update my paper to reference theirs.
21 May, 2024 at 9:04 am
Anonymous
That’s certainly amazing.
21 May, 2024 at 3:56 pm
Anonymous
Why the case of odd 𝑘 ≥ 5? see above
22 May, 2024 at 9:58 am
Terence Tao
I am not sure exactly what the thrust of your question is, but one can get significantly stronger upper bounds in the even case because one can exploit the birthday paradox: if one can find two separate -tuples in whose products have the same square-free part, then the product of the entire -tuple will be a square. One can use this trick to generate even tuples that multiply to a square, but not odd tuples. Similarly, it is easier to generate strong lower bounds in the odd case than in the even case, because the set of numbers that have an odd number of prime factors in a given set of primes, will necessarily have all odd products being a non-square, and all known counterexamples are based on this construction with a suitably optimized choice of .
23 May, 2024 at 3:19 am
Will Sawin
Lovely!
If one were to make the upper bound on the density obtained by this method explicit, it would go to 1 with k. Indeed it couldn’t possibly be higher than 1-1/k since we need to lower bound the probability that a k-tuple of random variables lies in A based only on information in the marginal distribution, requiring us to use the union bound, which loses at least 1 minus the density for each element of the tuple.
Might it be possible to prove a bound uniform in k at least 5 by some modification of this method? It seems tricky as one would have to prove the events n_i in A approximately independent in some sense. But actually I’m not sure that this is the main obstacle – it’s possible the current bounds from an explicit version of your argument go to 1 significantly faster than this.
23 May, 2024 at 9:40 am
Terence Tao
It seems one can get a lower bound on the leading constant that is uniform in by leveraging the even theory, which among other things tells us that any set of positive density will contain distinct elements multiplying to a square if is fixed and even. So then if is odd, and is a subset of that has density close to , then one can first locate a tuple of five elements multiplying to a square, delete those elements, and then locate a tuple of (which is an even number that is at least 4) elements that multiply to a square in the remaining set. (In the notation of the paper, this argument shows that for any fixed .) I’ll add this as a remark to the paper.
25 May, 2024 at 12:17 am
Csaba Sandor
Let be a positive integer. Then . Let be a largest subset of that does not contain distinct elements whose product is a square. It is known that . Then one can locate elements , multiplying to a square. Then is a set that does not contain distinct elements whose product is a square.
25 May, 2024 at 7:06 am
Terence Tao
Nice! So now the monotonicity of for odd is settled.
3 June, 2024 at 2:27 am
Anonymous
can you add it to your article (with ack. for Sandor of course)?
3 June, 2024 at 2:42 am
Anonymous
my bad, v2 already is there…
25 May, 2024 at 7:31 am
Terence Tao
One way to think about this phenomenon is that the set is not an “irreducible” set in some arithmetic geometry sense, but instead has multiple interesting “components”. In my paper I rely more or less exclusively on the component which has the parameterization for some natural numbers with . But there are other components as well; for instance one can retain this sort of parameterization just for , but have an unrelated parameterization , for the final two components; this is related to the monotonicity of for odd discussed previously. It is not clear a priori which components give the strongest bounds on (and perhaps the best bounds could even come from a synergy between multiple components, rather than just taking the best bound provided by a single component).
25 May, 2024 at 8:22 am
Terence Tao
As an experiment, I asked ChatGPT to provide Python code to compute the function , defined as the size of the largest subset of with no odd number of them multiplying to a square. This is a lower bound for all the for odd ; the quantity is also the minimal size of where ranges over -valued completely multiplicative functions. As it turned out the code ran flawlessly (other than I had to figure out install the sympy package, and how to print the results), and only took me a few minutes to complete: https://chatgpt.com/share/e021a5fb-5040-427b-a1a8-260fb31eeb08 . On consulting the OEIS, I found that the sequence already appears as https://oeis.org/A360659 (which also provided confirmation of the correctness of the GPT-provided code, which I also spot-checked with small values of ), although is not, but I just submitted it to the OEIS. (I haven’t tried the analogous exercise with the , which are harder to compute for , but presumably one could get GPT to produce some algorithm that could produce enough entries to check the OEIS.)
UPDATE: F(N) is now at https://oeis.org/A373114
25 May, 2024 at 9:55 am
Terence Tao
Indeed, ChatGPT could compute by brute force for up to 10 in less than a minute: https://chatgpt.com/share/79ab37e7-d14d-4ae7-96d6-a5794a0a9581 . I could verify and extend the sequence by hand to
and it does not appear to be in the OEIS; I will add it after the sequence is approved (this is currently in process). [ and already appear in the OEIS as https://oeis.org/A028391 and https://oeis.org/A013928 respectively.]
UPDATE: is now at https://oeis.org/A372306. is at https://oeis.org/A373119, is at https://oeis.org/A373178, and is at https://oeis.org/A373195.
25 May, 2024 at 3:55 pm
Anonymous
has the oeis led to discoveries?
26 May, 2024 at 3:57 pm
Terence Tao
There are over 10,000 research papers that cite the OEIS, so I presume some non-trivial fraction of these found it useful.
30 May, 2024 at 1:58 pm
Anonymous
Are there similar results for representations of cubes?
[Yes; see the paper https://arxiv.org/abs/2405.12088 referenced in a previous comment -T.]
9 June, 2024 at 3:57 am
Anonymous
Is it possible to show that the sequence is monotonic?
14 June, 2024 at 2:41 pm
Anonymous
Yes.
14 June, 2024 at 3:16 pm
Terence Tao
The are known to equal 1 for even , and are at most for odd , so the sequence is not monotonic over the full natual numbers; however they are monotonic non-decreasing for odd (see the comment of Sandor here).