A Set of Conjectures: Sequels to Fermat’s Last Theorem?

This story began yesterday, with this blog-post: https://robertlovespi.wordpress.com/2014/12/10/pythagorean-and-fermatian-triples-and-quadruples/ — but it hasn’t ended there. When discussing this with my wife (who, like myself, is also a teacher of mathematics) while writing that post, she speculated that more interesting things might happen — such as a “no solutions” situation, as is the case with Fermat’s Last Theorem — with a search for a Fermatian quadruple, if the exponent used were larger than three, the exponent I checked yesterday.

Tonight, therefore, I modified the program I used for the last post on this subject. Instead of searching for whole-number solutions to an + bn + cn = dwith n = 3, as I did yesterday, I looked for solutions to a4 + b4 + c4 = d4. As I did yesterday, I started with a search of all possibilities with numbers from 1 to 10, and was unsurprised when that quick, preliminary search yielded no solutions. I then ran the program again, but used it to search all possibilities using numbers for a, b, c, and d from 1 to 100. This took a while, for, with loops nested four deep, my computer had to check 1004 = 100 000 000 possibilities. The results are — tentatively — exciting, for this search, indeed, yielded no solutions, which is reminiscent of Fermat’s Last Theorem:

fermatian quadruple search of order four

We are now wondering if Fermat’s Last Theorem can be generalized indefinitely. Andrew J. Wiles proved that an + bn = cn has no solutions if n > 2. I’ve now written a program, and checked, and know that an + bn + cn = dn  has no solutions, with n = 4, for values of a, b, c, and d up to 100.

Could it be that an + bn + cn = dhas no solutions for any value of n > 3 — like Fermat’s Last Theorem, but with one more term added, and the exponent simply bumped up one place? If that is true, then, might it also be true that an + bn + cn + dn = ehas no solutions if n > 4? Might it be possible to extend this idea indefinitely, so that, with an equation containing k terms added together, to equal a single term, there are no solutions if n > k?

I know this much, at this point:

  1. I can either find a counterexample, to disprove one of these conjectured “sequels” to Fermat’s Last Theorem, if that counterexample is of reasonable size, provided a “smallish” counterexample actually exists, or
  2. With the assistance of friends of mine whose ability with mathematics, and computer programming, exceeds my own, we can extend this search for counterexamples to much higher limits, or
  3. This set of conjectures is, in fact, true, in which case we will find no counterexamples — and, if that is the case, I’m going to need to find some major-league help for this problem, for, well, if true, it’s going to be one monster of a job to prove it is true, especially in the general, unlimited form.
  4. Finally, I know that the prospect of playing any role, whatsoever, in extending Fermat’s Last Theorem to new levels is tremendously exciting.

I’m looking forward to seeing where this goes.

[Update: I’d like to thank my friend Andrew for finding the answer to this puzzle for me. Counterexamples have indeed been found for the four-term and five-term cases, one of which is 26824404 + 153656394 + 187967604 = 206156734. For six terms or more, this remains an unsolved problem. For more information related to this, please visit https://en.wikipedia.org/wiki/Euler’s_sum_of_powers_conjecture.]

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5 Responses to A Set of Conjectures: Sequels to Fermat’s Last Theorem?

  1. Pingback: Pythagorean and Fermatian Triples and Quadruples | RobertLovesPi

  2. You needn’t vary d from 1 to n. You can have the program calculate the fourth root of a^4 + b^4 + c^4, truncate the result, and just check if the resulting integer is a direct match. (Might double-check that this works right by applying the same idea to the simpler case of the Pythagorean theorem.) I guess it would be easy to try a couple of other small values of n and the case with 4 terms on the left.

    Liked by 1 person

  3. Another way to cut down your search:

    Let b vary from a to n
    Let c vary from b to n
    Let a start at 2
    (a = 5, b = 7, c =12 is no different from a = 7, b = 12, c = 5.)

    Liked by 1 person

    • Both good ideas. Thank you. I’m also calling on several friends on Facebook, asking for their help with this. I have a hunch it will soon become a group project. I also have a very strong feeling that the conjecture is true, but I don’t even know where to begin with the writing of a proof of any of this.


  4. Counterexamples found! At least for four and five terms, this conjecture, first posed by Euler, does not hold. Please see the update at the end of the post for more information.


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