Timeline for The new shortest open cubic equations
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| Oct 10 at 11:19 | history | edited | Bogdan Grechuk | CC BY-SA 4.0 |
Noted that equation (3) has been solved, and equation (4) is now the shortest open.
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| Oct 10 at 11:10 | answer | added | Bogdan Grechuk | timeline score: 3 | |
| Oct 9 at 5:48 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed some typos
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| Oct 8 at 21:16 | history | edited | Daniel Asimov | CC BY-SA 4.0 |
questions —> question, spelling
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| Oct 8 at 9:48 | comment | added | Bogdan Grechuk | Thanks, corrected. In the related equation, coefficient 10 is missing, so to solve (2) we need to find a solution to the related equation with z divisible by 10. | |
| Oct 8 at 9:47 | history | edited | Bogdan Grechuk | CC BY-SA 4.0 |
Corrected equation related to (2)
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| Oct 8 at 7:48 | comment | added | Gerry Myerson | The equation you say is related to (2) seems to be identical to (2). | |
| Oct 8 at 4:10 | comment | added | Dmitry Ezhov | @BogdanGrechuk. Infinite ascent can be easy interrupted at any step. Task is to do this in such a way as to find integer $z_i$. | |
| Oct 7 at 20:21 | comment | added | Bogdan Grechuk | Hm, sorry, if $z_i$ are not integers, then what is the meaning of these formulas? For me, the whole point is to find $x_i$ and $y_i$ such that $z_i$ are integers... Or these is some other meaning that I am missing? | |
| Oct 7 at 16:19 | comment | added | Dmitry Ezhov | Equation $-2+3x^3+y^2+xyz=0$ has infinite ascent over integers $x_i,y_i$. If $(x_0,y_0,z_0)$ is solution of this equation, then $$x_1= (-2 + y_0^2)/x_0, y_1= (2^2 3^2 + 3x_1^3)/y_0$$ $$x_2= (-2^3 3^4 + y_1^2)/x_1, y_2= (-2^7 3^{12} + 3x_2^3)/y_1$$ $$x_3= (-2^{11} 3^{20} + y_2^2)/x_2, y_3= (2^{26} 3^{50} + 3x_3^3)/y_2$$ $$...$$ But it didn't help to find ascent formulas with integer $z_i$. | |
| Oct 7 at 15:04 | comment | added | Bogdan Grechuk | Fedor Petrov. Equation $xy(x+y)=7z^2 + 1$ is an example of three-variable cubic equation with no solutions - this was proved by Michael Stoll using Jacoby symbol, see mathoverflow.net/questions/420896 . Equation $z(y^2-4)=x^3-x^2-2x+1$ is 3-variable, cubic, linear in $z$, and has no solutions, see mathoverflow.net/questions/417804 . If you ask about open equations, then equation $7x^3+2y^3=3z^2+1$ is 3-variable. cubic, open, and strongly conjectured to have no solutions, see mathoverflow.net/questions/467988 | |
| Oct 7 at 14:52 | comment | added | Fedor Petrov | What are examples of equations of this type (in 3 variables) which do not have solutions? | |
| Oct 7 at 14:50 | comment | added | Bogdan Grechuk | Sam Hopkins: I could equivalently ask one fixed non-moving question "What is the shortest cubic equation (if any exists) for which the problem of existence of integer solutions is independent from ZFC", and then exactly the same discussion would mean eliminating candidate answers to this fixed question. | |
| Oct 7 at 14:45 | comment | added | Sam Hopkins | I'm not sure how I feel about these "moving goalpost" type questions... | |
| Oct 7 at 14:41 | history | edited | Bogdan Grechuk | CC BY-SA 4.0 |
Updape in response to solution to (2) reported in comment. The next equation (3) added.
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| Oct 2 at 2:45 | comment | added | Dmitry Ezhov |
For equation $-2 - x + x^3 + y^2 + 10 x y z=0$ solution $(x,y,z)$=(23499130751021842,252697047241468990409432149765008357514,-1075346360334969622883)
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| Oct 1 at 17:51 | comment | added | Dmitry Ezhov | For equation $-2-x+x^3+y^2+xyz=0$ ascent also is work. $$x_1= (y_0^2-2)/x_0, y_1= (4-x_1^2+x_1^3)/(2*y_0)$$ $$x_2= (y_1^2-2)/x_1, y_2= (-2-x_2+x_2^3)/y_1$$ $$z_2= -(-2-x_2+x_2^3+y_2^2)/(x_2*y_2)$$ Then $(x_2,y_2,z_2)$ is new integer solution. | |
| Oct 1 at 7:55 | comment | added | Bogdan Grechuk | Now tested $y^2+10xyz+x^3-x-2=0$ up to $|x|\leq 10^9$ and $|y|\leq 10^9$. No solutions found. | |
| Sep 26 at 8:11 | comment | added | Bogdan Grechuk | If anyone checked new equation up to some large bound but found no solutions, please report the region checked to avoid duplicate work. | |
| Sep 20 at 13:04 | comment | added | Bogdan Grechuk | Ah, in fact heuristic works: constant positive probability of solutions between N and 2N exactly correcponds to the logarithmic number of solutions. If this "constant positive probability" is small, then we might need to search deep to find the first solution. | |
| Sep 20 at 10:24 | comment | added | Bogdan Grechuk | Not sure why this heuristic does not work, but computations show that for equations of this type the number of solutions is logarithmic, and in many examples the smallest known solution is huge, see e.g. mathoverflow.net/questions/466362 for equation $1 + x^2 + x^3 + y^2 + 9 x y z = 0$ | |
| Sep 20 at 9:39 | comment | added | Fedor Petrov | Heuristically, if $x$ is of order $N$ and $z$ of order $\sqrt{N}$ the value of $-x^3+x+2+25x^2z^2$ should be a perfect square with a constant positive probability. Thus, if no solution exist, this is not a coincidence (like for high degree equations) but must have clever reasons | |
| Sep 20 at 8:46 | comment | added | Bogdan Grechuk | Thank you, but the problem was whether there exist any solution, so first solution completely solves the problem for this equation. Instead of looking for further solution, it is more interesting to investigate the next-shortest open cubic equations. The question is updated to reflect this. | |
| Sep 20 at 8:44 | history | edited | Bogdan Grechuk | CC BY-SA 4.0 |
Question updated to say that the first equation has been solved by Dmitry Ezhov. The second equation is added.
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| Sep 19 at 18:08 | comment | added | Dmitry Ezhov |
$(x,y,z)$=(-3087382999, 74759753414, 6218157870)
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| Sep 15 at 9:51 | comment | added | Dmitry Ezhov | code pari/gp: my(x,d,D,a,b,z2,z,y); parfor(j=1, 10^8, x= 7+12*j; d= 3*x^4-x; D= divisors(d); for(i=1, #D, a= D[i]; b= d/a; z2= (a+b)/6; if(z2==floor(z2), if(issquare(z2), z= sqrtint(z2); y= a/x; /*y= (-b+6*z2)/x;*/ if(1 - 3*x^3 - xy^2 + 6*yz^2==0, print("("x","y","z")") ) ) ) ) ) | |
| Sep 15 at 4:45 | comment | added | Bogdan Grechuk | Thank you. Because x=12t+5, t=-84788477. Have you just tried all values of t up to this size or used any smarter method? | |
| Sep 14 at 13:40 | comment | added | Dmitry Ezhov |
$(x,y,z)$=(-1017461719,95574914,2350866170)
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| Mar 11 at 10:18 | history | asked | Bogdan Grechuk | CC BY-SA 4.0 |