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I do not have much formal mathematical education beyond high school, but I do like watching math YouTube videos. I recently read a comment on a video asking if anyone could prove that there are no positive integer solutions to the equation a2 + b2 + c2 - abc = 1. It took me a while, but I found a proof, and I then used a similar method to prove a more general claim.

For any given positive integer n, there exists a set of three positive integers a, b, and c satisfying the equation a2 + b2 + c2 - abc = n if and only if there exists such a set where a2 + b2 ≤ n. For example, if there are no positive integer solutions to the equation a2 + b2 + c2 - abc = 6 where a2 + b2 ≤ 6, then there are no positive integer solutions to that equation at all.

My proof is too long to paste into this post, so here is a link to a PDF of it: https://www.dropbox.com/scl/fi/9wsm5b4zsz9xpc0r7p0rk/naturalnumberproblem.pdf?rlkey=0gwqktvfu9215fskihlt2fvu3&dl=0

Since there are only finitely many pairs of positive integers who squares sum to a value less than or equal to a given number, you only need to check finitely many cases to determine whether there is a solution for a given n. For example, for n = 6, you only need to check {a = 1, b = 1} and {a = 1, b = 2}. Solving for c shows there are no positive integer solutions.

After discovering this, I wrote a program to that checked whether there were solutions for each n from 1 to 100. I was hoping to find a pattern in the numbers that have or don't have solutions, but I didn't find much. However, I did notice that about half of the numbers had solutions. That intrigued me, so I increased the range of my search to see if this percentage held.

Here are the results: From n = 100 to 999, the percentage of numbers with solutions stays between 49.3% and 52.5%. From n = 100 to 9999, the percentage of numbers with solutions stays between 49.8% and 50.4%. From n = 10000 to 99999, the percentage of numbers with solution stays between 50.1% and 50.7%

My conjecture is that this percentage stays around 50% and that the deviations get smaller and smaller over time. I assume there is some way to state this more formally using limits. Does anyone know if what I found has already been discovered and if my conjecture has already proven or disproven? What methods would you even use to do this?

As I said, I don't have much formal mathematical education. The math used in the proof linked above is fairly simple. Nothing more 'advanced' than modular arithmetic and proofs by contradiction. My intuition is that proving my conjecture would rely on math I've never even heard of, but maybe there is some (relatively) simple proof out there.

I can prove that ~41.6% of all numbers will not have solutions. (I am speaking loosely. I know this needs to be stated more formally with limits.) This is because there are no solutions to the equation when n is congruent to 3 mod 4, 3 mod 9, or 6 mod 9. This was verified by a computer trying every possible solution working from mod 2 to mod 800.

Let f(n) count the numbers less than or equal to n that are congruent to 3 mod 4, 3 mod 6, or 3 mod 9. There are no numbers that are congruent to both 3 mod 9 and 6 mod 9. The only numbers that are congruent to both 3 mod 4 and 3 mod 9 are those that are congruent to 3 mod 36, and the only one congruent to both 3 mod 4 and 6 mod 9 are those that congruent to 15 mod 36. Therefore, the limit of f(n)/n as n approaches infinity is equal to 1/4 + 1/9 + 1/9 - 1/36 - 1/36 = 5/12 = ~41.6%

One more thing I know (proof in that PDF) is that for all n except n = 2, there are either no positive integer solutions to the equation a2 + b2 + c2 - abc = n or there are infinitely many such solutions. For n = 2, there is only one solution {a = b = c = 1}. I doubt that is relevant to the conjecture, but it does not hurt to mention it.

One last thing: I'm tagging this as a number theory question and as a question about Diophantine equations. Let me know if that is incorrect.

EDIT #1: I want to add that I looked up the first dozen n with solutions in the OEIS and didn't find anything.

Here's a list of all n with solutions up to 100 in case it helps: 2, 4, 5, 8, 10, 13, 14, 17, 18, 20, 22, 25, 26, 28, 29, 32, 34, 37, 38, 40, 41, 44, 45, 49, 50, 52, 53, 54, 58, 61, 62, 64, 65, 68, 70, 72, 73, 74, 76, 77, 80, 82, 85, 88, 89, 90, 92, 94, 97, 98, 100

EDIT #2: My program just finished checking up to n = 1,000,000. From n = 100,000 to 1,000,000 the percentage of numbers with solutions stays between 50.7% and 51.6%. It's getting further from 50%, so I'm starting to doubt my conjecture is true.

If anyone wants to run the program themselves, here's a link to txt version of it:

https://www.dropbox.com/scl/fi/kb2p277sa6d29ieeedigk/mathprogram.txt?rlkey=sr5su9zny13y75qg0b5k1v0g4&dl=0

It's written in Python. I don't have any formal programming education either. Just started teaching myself Python a few weeks ago, so I apologize if it's confusing or inefficient code. Also, the program will not make much sense unless you've read the section in the linked proof called "A Note on Searching For Satisfactory Triplets."

EDIT 3: There was a typo in my post. I said there are no solutions when N is 3 mod 6. I meant to write 6 mod 9. Sorry!

EDIT 4: There were also a couple typos in my PDF where I used < when I meant to write ≤, but I have corrected them. (Unfortunately, this somehow created some new typos. It must be an issue with the conversion from .doc to .pdf but I don't think it effects the clarity much.)

EDIT 5: Some errors in the proof were fixed, and a new link has been created. This one should be viewable by anyone.

EDIT 6: Refined the approximation of how many numbers (less than N for sufficiently large N) don't have solutions from 40% to 41.6%.

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    $\begingroup$ yes, using Vieta Jumping, a solution to the original problem can be reduced to one that Hurwitz would have called a "ground solution." See zakuski.math.utsa.edu/~jagy/Hurwitz_A_1907.pdf It will take a while to see what I might have to contribute. $\endgroup$
    – Will Jagy
    Commented Aug 24, 2023 at 0:48
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    $\begingroup$ There is no such thing as "half of all natural numbers" in any meaningful sense. $\endgroup$ Commented Aug 24, 2023 at 1:19
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    $\begingroup$ @ThomasAndrews I think you're being too harsh on someone with little formal mathematical training. The question asked in the original post can readily and meaningfully be formalized by using limits and ratios: Let $P(m)$ be the number of $n \leq m$ such that the equation has a solution for $n$. The conjecture is $$\lim_{m \to \infty} \frac{P(m)}{m}=\frac 12.$$ $\endgroup$ Commented Aug 24, 2023 at 1:24
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    $\begingroup$ Make a list of the first few values of $n$ for which there s a solution, then type that sequence of values into oeis.org and see what comes up. $\endgroup$ Commented Aug 24, 2023 at 1:46
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    $\begingroup$ For all $n=a^2+b^2$, we have for all $c=ab$ that $a^2+b^2+c^2-abc=n$ $\endgroup$
    – Piquito
    Commented Aug 24, 2023 at 8:23

2 Answers 2

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This is not a complete answer.

Let $C$ be the set of $n$ you are interested in, i.e. those expressible as $x^2 + y^2 + z^2 - xyz$ for positive integers $x,y,z$.

Main Observation: $C$ is A351723 on OEIS, minus possibly some square numbers(for example $9$).

A351723 is defined as integers of the form $x^2 + y^2 + z^2 + xyz$, where $x, y, z \geq 0$.

Claim: Any positive integers $n \in C$ lies in A351723.

Proof: Suppose $n = x^2 + y^2 + z^2 - xyz$ where $x,y,z$ are positive integers. You have correctly shown that there is such a solution with $y^2 + z^2 \leq n$. Therefore, we can write $$n = x_1^2 + y^2 + z^2 + x_1 yz$$ with $x_1 = x - yz \geq 0$.

Similarly, one can prove that

Claim: Any non-square positive integer $n$ in A351723 lies in $C$.

Proof: If $n = x^2 + y^2 + z^2 + xyz$ and $n$ is not a square, then at least two of $x,y,z$ are non-zero. Say $y, z \neq 0$. Then we can write $$n = x_2^2 + y^2 + z^2 - x_2 yz$$ where $x_2 = x + yz$. Observe that $x_2, y, z$ are all positive integers.

In the comment section of A351723, Zhiwei Sun has conjectured that

Conjecture: If $a_n$ is the $n$-th term of A351723, then $a_n < 2n$.

which, if true, would imply your conjecture that the density of $C$ is at least $50\%$. So in a sense, you have recovered a conjecture made by a mathematician less than a year ago.

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    $\begingroup$ Thank you for bringing this to my attention. I noticed that the author of the sequence said that there are no solutions to his equation when n is congruent to 3 mod 4. I checked, and there are also no solutions when n is congruent to 3 mod 9 or 6 mod 9, just like my equation. (This is also follows from your proof that A351723 is C plus some square numbers and the fact that no square numbers are congruent to 3 mod 9 or 6 mod 9). Thus, the density of A351723 is also less no greater than 7/12 for reasons explained in my post. (Please correct me if I am using term "density" incorrectly.) $\endgroup$ Commented Aug 25, 2023 at 19:22
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I have been pointed to a reference that resolves this problem.

Ghosh, A., Sarnak, P. Integral points on Markoff type cubic surfaces. Invent. math. 229, 689–749 (2022)

Unfortunately, I don't have the technical expertise to understand the proof in this paper. Its main theorem is

Theorem: For $100\%$ of integers $k$ such that $k$ is not congruent to $3$ modulo $4$ or to $\pm 3$ modulo $9$, there exists integers $x,y,z$ such that $k = x^2 + y^2 + z^2 - xyz$.

Here by "$100\%$ of integers $k$", I mean the following statement. Let $A(N)$ be the number of integers $k \in [1, N]$ such that there exists integers $x,y,z$ with $k = x^2 + y^2 + z^2 - xyz$, and $B(N)$ be the number of integers $k \in [1, N]$ not congruent to $3$ modulo $4$ or to $\pm 3$ modulo $9$. Then $$\lim_{N \to \infty} \frac{A(N)}{B(N)} = 1.$$

Of course, the OP's question asks for positive integer solutions to this problem. My previous answer shows that if $k$ is not square and $k = x^2 + y^2 + z^2 - xyz$ has an integer solution, it also has a positive integer solution. So the conclusion is

Corollary: Exactly $7 / 12$-fraction of the integers $k$ can be expressed in the form $x^2 + y^2 +z^2 - xyz$ for positive integers $x,y,z$.

This affirmatively answers both the OP's conjecture and Zhiwei Sun's conjecture.

P.S. Ghosh and Sarnak are two of the greatest number theorists alive, and Inventiones is among the best journals in mathematics. This testifies to the exceptional quality of OP's conjecture.

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    $\begingroup$ I think a mistake was made somewhere. 46 is not a perfect square and is not congruent to 3 mod 4, 3 mod 9, or 6 mod 9. Thus, according to you, there should be a positive integer solution to the equation: x^2 + y^2 + z^2 − xyz = 46. According to my proof, there should then be a solution where x^2 + y^2 are less or equal to 46. However, I can find no such solution. I've checked all values of x and y where the sum of their squares is less than or equal to 46 and solved for z, but z is never a positive integer in these cases. $\endgroup$ Commented Nov 18, 2023 at 2:11
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    $\begingroup$ The theorem I quoted does not claim that $k$ can be expressed as $x^2 + y^2 + z^2 - xyz$ for all integers $k$ satisfying the congruences. Rather, it claims that the density of such integers approach $1$ as we consider larger and larger $k$. There are indeed be small counterexample (like the $46$ you found), but these become rarer and rarer. $\endgroup$
    – abacaba
    Commented Nov 18, 2023 at 2:40
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    $\begingroup$ I have clarified this in the answer. Thanks for pointing this out. $\endgroup$
    – abacaba
    Commented Nov 18, 2023 at 2:43

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