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:
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.
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/.]
Offhand I would expect there to be infinitely many quadruples, although I suspect it’s an annoying bit of work to actually generate them. Shall consider it, though.
Maybe Fermat was looking for the wrong thing! Since there are triples satisfying a^2 + b^2 = c^2 and quadruples satisfying a^3 + b^3 + c^3 = d^3, perhaps he needed to look for numbers that satisfy a^4 + b^4 + c^4 + d^4 = e^4, etc. Regardless, Robert, you may have come across something groundbreaking here.
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This isn’t groundbreaking, but it’s given me an idea for a new research project in number theory which might be. Maybe. We’ll see.
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I spent months at a time in the 90’s searching for a simple, elegant proof of Fermat’s theorem, algebraic or geometric, but couldn’t make it connect. I discovered a number of interesting corollaries, though.
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Please check the new sequel to this post — I need help with this new problem you’ll see described there, from anyone who is interested!