Solve 3 Integer Puzzle: a^3+b^3=17c^3

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Discussion Overview

The discussion revolves around the problem of finding three integers, a, b, and c, such that a^3 + b^3 = 17 * c^3. Participants explore various methods and theories related to this equation, including number theory, algebraic approaches, and computational searches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests starting with factoring a^3 + b^3 and notes that since 17 is prime, it must divide one of the factors, assuming a, b, and c have no common factors.
  • Another participant expresses interest in using a brute-force search in Mathematica to find integer solutions, questioning the existence of an algebraic method for this problem.
  • Concerns are raised about the trivial solutions (0,0,0) and (1,-1,0), with a participant noting that Mathematica has not found any valid instances of the equation being true.
  • One participant mentions using modular arithmetic, specifically mod 8, to derive conditions on a, b, and c, indicating specific cubic remainders that must hold.
  • Another participant discusses findings related to mod 7, suggesting that 7 divides c and that a^3 and b^3 are ±1 mod 7, while also noting implications mod 17.
  • There is a mention of the computational limits of Mathematica, with one participant expressing a desire to run more extensive searches for solutions and questioning the existence of a proof for no solutions.
  • Some participants speculate about the relationship of the problem to Fermat's theorem and the potential for known results in cubic number fields.
  • One participant expresses uncertainty about their knowledge in number theory while still showing interest in the topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of solutions to the equation. There are multiple competing views regarding the methods to approach the problem and the implications of modular arithmetic. The discussion remains unresolved.

Contextual Notes

Participants mention limitations in their computational resources and knowledge of number theory, which may affect their ability to explore the problem fully. There are references to trivial solutions and the need to consider negative integers, as well as the potential for existing knowledge in cubic number fields.

IndustriaL
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I want to find three integers, a b and c, with a^3 + b^3 = 17 * c^3.
all each the smallest integer possible
 
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you could factor a^3+b^3 for a start, I suppose. 17 being prime must divide one of the factors. we can also suppose that a,b a nd c have no common factors.
 
Well I like Number Theory too. Before I start coding this into Mathematica in a brute-force search, you know, (1,1), (1,2),. . . (2,1),(2,2), . . . I suppose there is no algebraic method for doing so? And while I'm at it, why not just search for the relation involving an arbitrary prime instead of 17 and then see if any conclusion can be drawn say for the first 100 primes.
 
Assuming your not looking for the trivial solution (0,0,0) and other trivial solutions like (1, -1, 0). Then mathematica can't find a single instance where it is true.
 
there are algebraic ways of factoring such things in number fields, though i don'y know much about them - they are of elementary (in the sense of important) interest in algebraic number theory.

you can also get conditions on a,b and c by considering it modulo some inetger. eg doing so mod 8 the cubic remainders are 0,1,3,5,7 and we need

a^3+b^3=c^3 mod 8

so that the only non trivisl way to do it is to have a^3=1,b^3=7 mod 8 or a^3=3, b^3=5 mod 8 or a^3=b^3=c^3=0 mod 8
 
Last edited:
Zurtex said:
Assuming your not looking for the trivial solution (0,0,0) and other trivial solutions like (1, -1, 0). Then mathematica can't find a single instance where it is true.

Why not Zurtex? I respect you and Matt's opinions but I'm stubborn. You mind if I check?
 
perhaps zurtex checked it in mathematica?

i meant to say, that the case a^3=b^3=c^3 can be dismissed as trivial since it implies that all of a,b,c are even thus it is not a "primitive" solution, but i forgot the cases

a^3=0 and b^3=c^3 mod 8
 
looking mod 7 i think we find that 7 divides c and a^3 and b^3 are +/-1 mod 7 and mod 17 that a+b=0
 
Alright, I'd like to qualify Zurtx's statement if I may: Mathematica cannot find any value of a and b under 5000 which satisfy the equation. Frankly, if I had access to a faster PC I'd run it up to a million at least as well as optimize my algorithm. It kinds looks like it's related to Fermat's theorem. Is there a proof that there is no solution?

Wait, he did say integers. Need to check some negative numbers too. Didn't say I was good in Number Theory, only that I'm interested in it. :smile:
 
Last edited:
  • #10
saltydog said:
Alright, I'd like to qualify Zurtx's statement if I may: Mathematica cannot find any value of a and b under 5000 which satisfy the equation. Frankly, if I had access to a faster PC I'd run it up to a million at least as well as optimize my algorithm. It kinds looks like it's related to Fermat's theorem. Is there a proof that there is no solution?
Actually I just used the FindInstance function for c > 0 and then for c < 0.
 
  • #11
Zurtex said:
Actually I just used the FindInstance function for c > 0 and then for c < 0.

FindInstance? Suppose that's a 5.0 feature. I have 4.1.

IndustriaL, where you got this anyway? I'd like to see some closure in the matter. You know Richard Guy has a nice book out, "Unsolved Problems in Number Theory". Don't suppose anything like this is in there?
 
  • #12
i suspect if you have a basic knowledge of cubic number fields then the answer is known - we can do it for quadratics in quadratic number fields.
 
  • #13
matt grime said:
i suspect if you have a basic knowledge of cubic number fields then the answer is known - we can do it for quadratics in quadratic number fields.

Very well Matt. I'll leave it there.
 
  • #14
i say that but i am only guessing from the results about when we can solve x^2-dy^2=something; i don't have any knowledge of this at all.
 

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