Timeline for Integer solution to $(3x-1)y^2 + x z^2 = x^3-2$
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| 3 hours ago | comment | added | Tong Lingling | Thank you. From Python computation, the values of $x$ among positive integers up to $1000$ for which the equation has rational solutions are: $$x\in \{18, 64, 151, 199, 376, 391, 448, 504, 738, 898, 904, 967\}$$. There are $90$ such $x$ below $10000$. | |
| 15 hours ago | comment | added | Bogdan Grechuk | These days such questions are answered instantly by ChatGPT - for example, take x=−8,y=−8/5,z=15/2 | |
| 19 hours ago | comment | added | Tong Lingling | Does there exist a positive integer $x$ such that it has a rational solution ? | |
| Mar 16 at 14:20 | comment | added | Aurel | Oh, I see I should have been more careful. Depending on the density of $p$'s the decay is not $c/\log(X)$ but $\log(X)^{-c}$. So the prediction would be a power of log, not log log. I should work out the actual power and consider running the large computation, but I don't have time at the moment. | |
| Mar 15 at 7:13 | history | edited | Bogdan Grechuk | CC BY-SA 4.0 |
added 22 characters in body
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| Mar 14 at 18:30 | comment | added | Bogdan Grechuk | Now checked for |y| up to three billions ($3\cdot 10^9), still no solutions. I will stop search here, at least temporary. | |
| Mar 13 at 19:41 | comment | added | Bogdan Grechuk | Is this heuristic about the same for equations $(3x-1)y^2 + x z^2 = x^3+a$ for other values of $a$, or $a=-2$ is very special? For example, I just looked for solutions of 12 equations with $a=\pm 2, \pm 3, \dots, \pm 7$ in the range $10^5 < |y| < 8\cdot 10^5$, and found that these equations combined have 15 solutions in this range (up to the sign changes in $y$ and $z$), with exactly 5 solutions in each range $10^5 < |y| < 2\cdot 10^5$, $2 \cdot 10^5 < |y| < 4\cdot 10^5$, and $4 \cdot 10^5 < |y| < 8\cdot 10^5$. This clearly suggests $\log X$ decay, as originally suggested. | |
| Mar 13 at 19:24 | comment | added | Aurel | ... a sum of $1/x\log(x)$, which diverges, but only like $\log\log X$. If that is right, the solutions will exist but be very large, and inaccessible by simply enumerating the possible $x$ (which I was considering trying in parallel). | |
| Mar 13 at 19:23 | comment | added | Aurel | @WillSawin, BogdanGrechuk: Actually I am now convinced that the number of prime factors does not change WillSawin's heuristic. However there is another partial obstruction that reduces the possible $x$. For every $p$ such that $2$ is not a square mod $p$ (this is a positive proportion of primes), $x$ cannot be $0$ modulo $p$. (there is a similar obstruction for $3x-1$ and for $x^3-2$) In the heuristic this reduces the proportion of candidate $x\le X$ by $\prod_{p<X}(1-1/p)$ where the product is over the primes with the obstruction, and this decays like $c/\log(X)$. So the heuristic becomes... | |
| Mar 13 at 12:43 | comment | added | Thomas Browning | Here's one way to look for huge solutions: write the equation as $x(9z^2+53)=(3x-1)(3x^2+x+18-9y^2)$ so that $9y^2=3x^2-(m-1)x+18$ and $9z^2=m(3x-1)-53$. Then $m=3k+1$, so $3y^2=x^2-kx+6$ and $3z^2=(3k+1)x-(k+18)$. Now run through values of $k$ for which these equations have local solutions (e.g., this already imposes $k\equiv7\pmod{24}$), solve the Pell equation $(2x-k)^2-12y^2=k^2-24$ (i.e., for each fundamental solution, multiply by powers of fundamental unit), and check whether any of these satisfy $3z^2=(3k+1)x-(k+18)$. I've checked up to $k=10^7$ this way (assuming my code is right). | |
| Mar 13 at 7:45 | comment | added | Bogdan Grechuk | So, what is the prediction then? Logarithmic number of solutions or "a convergent sum" predicting the non-existence of large solutions at all? | |
| Mar 13 at 1:08 | comment | added | Will Sawin | @Aurel Yes, indeed. The class number being $x$ on average should come from moments of $L$-functions heuristics, which predict that $L(1, \left( \frac{\cdot}{ x (3x-1} )$ has a constant average value. | |
| Mar 12 at 23:07 | comment | added | Aurel | Ah but when there is a square root modulo each $p$ then there is a large total number of square roots modulo their product. | |
| Mar 12 at 23:05 | comment | added | Aurel | @WillSawin Where do you get the $O(1/x)$ probability? The discriminant is about $x^2$ so the class number should be about $x$ but could vary by a log factor. In addition, for each prime factor $p$ of $x^3-2$ that occurs with odd valuation, $x(3x-1)$ has to be a square; there are about $\log\log x$ such $p$, giving a heuristic probability of $1/(\log x)^{\log 2}$. So the convergence of the sum over $x$ very much depends on the log factors, and if we take $x$ for an estimate of the class number it gives a convergent sum, doesn"t it? | |
| Mar 11 at 6:16 | comment | added | Bogdan Grechuk | Now checked for $|y|$ up to billion ($10^9$), still no solutions. | |
| Mar 11 at 5:04 | comment | added | Tomita | $y^2=2+kx=48a^4+144a^2+110$ has no integer solution. | |
| Mar 11 at 1:19 | comment | added | Dmitry Ezhov | For $x^2-3kx+k-6=z^2$ exist $x = 18 + 12 a^2, k = 6 + 4 a^2, z = 2 a$, where $a \in \mathbb{Z}$ | |
| Mar 10 at 21:47 | comment | added | Bogdan Grechuk | What I am doing is try $y$ in the correct residue class in order, for each $y$ try all divisors $x$ of $y^2-2$, compute $k=(y^2-2)/x$, and check if $x^2-3kx+k-6$ is a perfect square. If we assume that the smallest solution is huge, say $20$ digits or more, we never find it this way. What else we can do? Of course, we may try $k$ in order, for fixed $k$ find $(x,z)$ from the first equation, which can be huge. But then if $x$ is huge, what is the chance that $2+kx$ happen to be a perfect square? So, I do not see a good way to look for huge solutions. | |
| Mar 10 at 21:26 | comment | added | Guruprasad | $y^2$ is Cong 2 mod $x$ so take $y^2=2+kx$ and substitute $y^2$ in original equation and after simplification we get, $ x^2-3kx +k-6=z^2$ and $y^2=2+kx$ . solve first equation for $x$ and $z$ | |
| Mar 10 at 18:38 | comment | added | Will Sawin | Since $3x-1$ and $x$ have to have the same sign, the equation is positive definite in $y,z$ and so $y$ and $z$ both must be $O(x)$. The heuristic probability that a solution exists for fixed $x$ is going to be $O(1/x)$, and then unless the constant in the big $O$ is very large, summing only over $x$ congruent to $34$ mod $48$ is going to give a prediction for the number of solutions with $x<X $ that is a small constant times $\log X$. So maybe the search just isn't big enough. | |
| Mar 10 at 13:50 | comment | added | Bogdan Grechuk | Yes, moreover, modulo $16$ analysis return only two even values for $x$, $2$ and $6$. Because all prime divisors of $x/2$ are $\pm 1$ mod $8$, $6$ is impossible. Further, modulo $9$ analysis shows that $x$ is $1$ mod $3$. Hence, $x$ is $34$ mod $48$. | |
| Mar 10 at 10:44 | comment | added | Fedor Petrov | For what it worth, $x$ can not be odd: we have $y^2\equiv 2\pmod x$, thus all prime divisors of $x$ are $\pm 1 \pmod 8$, thus so is $x$, but both cases are impossible modulo 8 | |
| Mar 10 at 9:48 | comment | added | Chris Wuthrich | This affine varietiy has rational points, like $(4/3,1/3,1/6)$ which, presumably, makes Brauer-Manin obstruction "stuff" harder to use. | |
| Mar 10 at 8:11 | comment | added | Bogdan Grechuk | @David Roberts - yes, this is because of $Ay^2+Bz^2=C$, and because I decided that this is the nicest-looking way to write the equation. You may suggest your own, better way :) | |
| Mar 10 at 8:09 | comment | added | Bogdan Grechuk | @Fedor Petrov I am not able to check Brauer-Manin obstruction by hand, and, as far as I know, currently there are no ready to use Magma code that can check this. | |
| Mar 10 at 6:59 | comment | added | David Roberts♦ | May I ask why it's written in the way it is? Is that so that for fixed $x$ you have $Ay^2 + Bz^2 = C$? | |
| Mar 10 at 6:31 | comment | added | Fedor Petrov | What about Brauer-Manin obstruction stuff? | |
| Mar 9 at 22:54 | history | asked | Bogdan Grechuk | CC BY-SA 4.0 |