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. -----
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