Re: [math-fun] 33 as sum of 3 integer cubes
What about other residue class restrictions besides mod 9 ? Googling, I found this discussion, <https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as-a-sum-of-three-cubes-in-infinitely-many-ways>, where someone states no other such restrictions are known. Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation x^3 + y^3 + z^3 = N has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case. Another cute tidbit: N = 3 has two known solutions mod permutations: 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known. —Dan Allan Wechsler wrote: ----- Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is x^3 + y^3 + z^3 = 42. -----
Oh, for some reason, I thought that once you had one solution there was known technology for finding an infinite sequence of them. On Sat, Mar 9, 2019 at 3:45 PM Dan Asimov <dasimov@earthlink.net> wrote:
What about other residue class restrictions besides mod 9 ?
Googling, I found this discussion, < https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as-a-sum-of-three-cubes-in-infinitely-many-ways>, where someone states no other such restrictions are known.
Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation
x^3 + y^3 + z^3 = N
has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case.
Another cute tidbit: N = 3 has two known solutions mod permutations:
1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known.
—Dan
Allan Wechsler wrote: -----
Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is
x^3 + y^3 + z^3 = 42. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Regarding Dan's query, 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? I there are more for 3, but I don't know. But there are lots similar to this, for example: (-95)^3 + 91^3 + 47^3 = (-77)^3 + 76^3 + 26^3 = 19^3 +( -16)^3 +(-14)^3 = 3^3 + (-2)^3 + 0^3 = 19. On Sat, Mar 9, 2019 at 2:45 PM Dan Asimov <dasimov@earthlink.net> wrote:
What about other residue class restrictions besides mod 9 ?
Googling, I found this discussion, < https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as-a-sum-of-three-cubes-in-infinitely-many-ways>, where someone states no other such restrictions are known.
Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation
x^3 + y^3 + z^3 = N
has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case.
Another cute tidbit: N = 3 has two known solutions mod permutations:
1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known.
—Dan
Allan Wechsler wrote: -----
Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is
x^3 + y^3 + z^3 = 42. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Regarding Dan's query, 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? I doubt there are more for 3, but I don't know. But there are lots similar to this, for example: (-95)^3 + 91^3 + 47^3 = (-77)^3 + 76^3 + 26^3 = 19^3 +( -16)^3 +(-14)^3 = 3^3 + (-2)^3 + 0^3 = 19. On Sat, Mar 9, 2019 at 2:45 PM Dan Asimov <dasimov@earthlink.net> wrote:
What about other residue class restrictions besides mod 9 ?
Googling, I found this discussion, < https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as-a-sum-of-three-cubes-in-infinitely-many-ways>, where someone states no other such restrictions are known.
Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation
x^3 + y^3 + z^3 = N
has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case.
Another cute tidbit: N = 3 has two known solutions mod permutations:
1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known.
—Dan
Allan Wechsler wrote: -----
Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is
x^3 + y^3 + z^3 = 42. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I'm interested in the origin of James's doubt. I have an intuition that if the sum is not 9n +/- 4, then there ought to be an infinite number of solutions. There must be a probabilistic argument, but I can't do the calculus in my head... On Sat, Mar 9, 2019 at 9:03 PM James Buddenhagen <jbuddenh@gmail.com> wrote:
Regarding Dan's query, 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? I doubt there are more for 3, but I don't know. But there are lots similar to this, for example: (-95)^3 + 91^3 + 47^3 = (-77)^3 + 76^3 + 26^3 = 19^3 +( -16)^3 +(-14)^3 = 3^3 + (-2)^3 + 0^3 = 19.
On Sat, Mar 9, 2019 at 2:45 PM Dan Asimov <dasimov@earthlink.net> wrote:
What about other residue class restrictions besides mod 9 ?
Googling, I found this discussion, <
https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as...
, where someone states no other such restrictions are known.
Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation
x^3 + y^3 + z^3 = N
has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case.
Another cute tidbit: N = 3 has two known solutions mod permutations:
1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known.
—Dan
Allan Wechsler wrote: -----
Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is
x^3 + y^3 + z^3 = 42. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
My doubt that there are probably not infinitely many integer solutions to x^3 + y^3 + z^3 = 3 is not from some great understanding of the problem, but just from having played with it for a couple of days on a computer. Rational yes, but integer not obvious. See also this paper: https://www.uni-math.gwdg.de/jahnel/Arbeiten/elk_ants6c.pdf On Sun, Mar 10, 2019 at 10:10 AM Allan Wechsler <acwacw@gmail.com> wrote:
I'm interested in the origin of James's doubt. I have an intuition that if the sum is not 9n +/- 4, then there ought to be an infinite number of solutions. There must be a probabilistic argument, but I can't do the calculus in my head...
On Sat, Mar 9, 2019 at 9:03 PM James Buddenhagen <jbuddenh@gmail.com> wrote:
Regarding Dan's query, 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? I doubt there are more for 3, but I don't know. But there are lots similar to this, for example: (-95)^3 + 91^3 + 47^3 = (-77)^3 + 76^3 + 26^3 = 19^3 +( -16)^3 +(-14)^3 = 3^3 + (-2)^3 + 0^3 = 19.
On Sat, Mar 9, 2019 at 2:45 PM Dan Asimov <dasimov@earthlink.net> wrote:
What about other residue class restrictions besides mod 9 ?
Googling, I found this discussion, <
https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as...
, where someone states no other such restrictions are known.
Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation
x^3 + y^3 + z^3 = N
has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case.
Another cute tidbit: N = 3 has two known solutions mod permutations:
1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known.
—Dan
Allan Wechsler wrote: -----
Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is
x^3 + y^3 + z^3 = 42. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Andrew Booker, Cracking the Problem with 33, https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf --Ed Pegg Jr On Sun, Mar 10, 2019 at 2:01 PM James Buddenhagen <jbuddenh@gmail.com> wrote:
My doubt that there are probably not infinitely many integer solutions to x^3 + y^3 + z^3 = 3 is not from some great understanding of the problem, but just from having played with it for a couple of days on a computer. Rational yes, but integer not obvious. See also this paper: https://www.uni-math.gwdg.de/jahnel/Arbeiten/elk_ants6c.pdf
On Sun, Mar 10, 2019 at 10:10 AM Allan Wechsler <acwacw@gmail.com> wrote:
I'm interested in the origin of James's doubt. I have an intuition that if the sum is not 9n +/- 4, then there ought to be an infinite number of solutions. There must be a probabilistic argument, but I can't do the calculus in my head...
On Sat, Mar 9, 2019 at 9:03 PM James Buddenhagen <jbuddenh@gmail.com> wrote:
Regarding Dan's query, 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? I doubt there are more for 3, but I don't know. But there are lots similar to this, for example: (-95)^3 + 91^3 + 47^3 = (-77)^3 + 76^3 + 26^3 = 19^3 +( -16)^3 +(-14)^3 = 3^3 + (-2)^3 + 0^3 = 19.
On Sat, Mar 9, 2019 at 2:45 PM Dan Asimov <dasimov@earthlink.net> wrote:
What about other residue class restrictions besides mod 9 ?
Googling, I found this discussion, <
https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as...
, where someone states no other such restrictions are known.
Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation
x^3 + y^3 + z^3 = N
has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case.
Another cute tidbit: N = 3 has two known solutions mod permutations:
1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known.
—Dan
Allan Wechsler wrote: -----
Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is
x^3 + y^3 + z^3 = 42. -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Allan Wechsler -
Dan Asimov -
Ed Pegg Jr -
James Buddenhagen