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.]

Pythagorean and Fermatian Triples and Quadruples

Pythagorean triples are familiar to almost everyone who has studied mathematics:  whole numbers which serve as solutions to the Pythagorean Theorem, a² + b² = c². Examples include 3, 4, 5; as well as 5, 12, 13; and 8, 15, 17; and 7, 24, 25. It has been proven that there are infinitely many Pythagorean triples.

Fermatian triples, on the other hand, don’t exist, which humanity finally found out, definitively, when Andrew J. Wiles finally proved Fermat’s Last Theorem, in the mid-1990s. If they did exist, they would satisfy an + bn = cn, with n > 2, and all numbers involved being whole numbers. This “only” took over 300 years to prove, and will forever stand as one of the greatest achievements in number theory.

This morning, while driving to work (one must think about something while driving, right?), I started trying to find Pythagorean quadruples:  sets of four whole numbers which satisfy a² + b² + c² = d², which can be pictured as pairs of solutions to the Pythagorean Theorem, for right triangles in which the hypotenuse of one triangle is a leg of the next, and the triangles exist in perpendicular planes. It didn’t take me long to figure out that 3, 4, 12, 13 is a Pythagorean quadruple, based on this mental image:

Pythagorean quadruple

The next logical step was to wonder . . . are there Fermatian quadruples? Those would be, of course, whole-number solutions to an + bn + cn = dn, with n > 2. However, I had to teach all day, and did not have the time, until after work, to indulge my curiosity on this subject.

Once the workday was over, I contemplated looking over lists of perfect cubes (since three as the exponent is the logical place to start looking), seeking three of them that would sum to a fourth, and quickly decided that was not the approach I wanted to use . . . because it sounded ridiculously tedious. Mathematics is supposed to be fun, after all, not an exercise in boredom. I therefore resolved to use a different approach, and wrote a short program, in BASIC, to check all sets of four numbers between one, and any number I chose, to seek exponent-three Fermatian quadruples. For the first trial run of this program, I considered checking all numbers between 1 and 100, but, since the program involves quadruple-nested loops, I decided I did not want to wait for my computer to check 100^4 = 100 million different combinations, so I made my first check with much more modest search parameters: only the numbers from one to ten. To my surprise, this search actually revealed two Fermatian quadruples.

fermatian quadruple

The two Fermatian quadruples this search revealed are 1, 6, 8, 9 (since 1 + 216 + 512 = 729); and 3, 4, 5, 6 (since 27 + 64 + 125 = 216). With a more extensive search, I could easily find more, and I suspect there are infinitely many of them, as is the case with Pythagorean triples . . . but this is enough recreational mathematics, I think, for one day.

[Later edit — to see what happened the next day, with this idea, just check this post: https://robertlovespi.wordpress.com/2014/12/11/a-set-of-conjectures-sequels-to-fermats-last-theorem/.]