Fermat's last theorem and Pythagorean quadruples

We have X^2 + y^2 = z^2…the Pythagorean theorem

examples of perfect Pythagorean triples are:

3:4:5
5:12:13

Fermat’s Last Theorem states for X^n + Y^n = Z^n, no perfect triple can be found for integer values of n larger than 2.

Messing around one day with my calculator, I have (re)discovered what I call Pythagorean quadruples, of the form X^3+Y^3+Z^3=W^3…examples:

3:4:5:6
7:14:17:20

My question is: can a perfect Pythagorean quadruple be found for
X^n+Y^n+Z^n=W^n with an integer value for n larger than 3?

Solutions exist for n=4

ETA: And n=5

Definitively answered.

Flawless victory.

One interesting point about the link Rysto gives is that the solution with the smallest integers is 95800^4 + 217519^4 + 414560^4 = 422481^4, which are fairly large values.

For years before FLT was proved, mathematicians kept coming up with proofs that any integer solution had to have values higher than some enourmous limit X, so one could be forgiven for thinking that FLT was in all likelyhood true even before Wiles proof came along. The example above justifies skepticism even in the face of overwhelmig finite evidence…

In David wells’nteresting book The Dictionary of Interesting Numbers there are examples of these for many higher values of n. I noticed that in each case, the number of terms added had to be equal to n, and so I thought “Aha! a pattern!” But for n = 5 or n = 6 it breaks dow, with vcounterexamples that require fewer than n terms.

A related story: A long while ago I was interested in whole number solutions to the Pythagorean equation (Pythagorean Triples, although I didn’t know there was a term for it at the time). Based on limited examples and some slightly flawed reasoning, I concluded that as you used larger and larger numbers (3,4,5/ 5,12,13/ 7,24.25 etc.) the only solutions you would find would represent triangles that were increasingly narrow slivers. Recently I found an excellent Ask Dr. Math page, in which I found I had been dead wrong all along. Part of the reason was you have to go up to 20 before you find a counter-example ( 20[sup]2[/sup] +21[sup]2[/sup] =29[sup]2[/sup]).

Provided “perfect” is in this case understood to mean “integer”.

is 3^3 + 4^3 + 5^3 = 6^3 a perfect triplet

Yes.

then is a^4 + b^4 + c^4 = d^4?

For exponent 3, there are infinitely many solutions in relatively prime integers (no common divisor). I don’t know about higher exponents.

3^3 + 4^3 + 5^3 = 6^3 reminds of the story that when Remanujan told Hardy that 1729 is the smallest positive integer that is the sum of two cubes in two different ways, he was wrong (if that’s what he said, which was Hardy’s story). 91 = 3^3 + 4^3 = (-5)^3 + 6^3 is the smallest. Moreover 8*91 = 728 is the sum of two cubes in three different ways, multiply the above by 2^3 to get two of them and then observe that 728 = (12)^3 + (-10)^3.

The phrase as given by Hardy was “smallest number expressible as the sum of two cubes in two different ways”.

I think it’s reasonably charitable (even assuming Hardy’s anecdotal retelling was to be word-for-word) to allow that “cube” is from context to be taken to mean “cube within the semiring of natural numbers”. One might just as well object that, for example, 9 can be written as both 1^3 + 2^3 and 1.5^3 + (cube root of 5.625)^3…

While thinking about some of the things in this thread (i.e. which integers can be sides of a ‘primitive’ Pythagorean triangle), I came across a wonderful site which might have been titled, “Everything You Ever Wanted to Know About Pythagorean Triangles”. Alas, it doesn’t even seem to have a name; but it is terrific. Worth a look if you’ve followed this thread.

Unnamed but wonderful Pythagorean Triangle site.