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For example, the triple

$(12, 16, 20)$

has these exponents:

$(12^1, 4^2, 20^1)$ or $(12^1, 2^4, 20^1)$

denoted by

$(1,2,1)$ or $(1,4,1)$


Here are more triples (ignore simple (1,1,1) cases):

fix-trips

I ran millions of triple tests looking for two identical exponents > 1, but found no matches.

Is there an existing proof or can someone prove it?

Thanks!

vengy
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    Have a look at Fermat's last theorem. – Donald Splutterwit Dec 10 '20 at 00:51
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    $a^n + b^n \ne c^n$ for integers $a,b,c,n$ and $n>2$ There is a proof of this, but I can't fit it in this comment box. – Doug M Dec 10 '20 at 00:55
  • What? 600 characters is plenty! ;) – vengy Dec 10 '20 at 00:57
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    @vengy Are you asking about all $3$ values having the same $\gt 1$ power, or just at least $2$ of them? For the first case, as the comments indicate, this is not possible. However, for the second case, that is something I'm not sure about offhand, but it seems like an interesting question to me. – John Omielan Dec 10 '20 at 00:57
  • Initially it was all 3 values the same, but if at least two are impossible too, that would amount to FLT, right? – vengy Dec 10 '20 at 01:06
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    vengy--- You have in your list at least some cases of 2 of 3 exponents same. So that is not impossible, so that case is not a case of FLT. – coffeemath Dec 10 '20 at 01:12
  • Yep, ignore the 1 exponents being the same. I was thinking about exponents > 2. – vengy Dec 10 '20 at 01:15
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    @vengy In your examples, note the first Pythagorean triplet $(3,4,5)$ should have exponents of $(1, 2, 1)$. Also, are you thinking of exponents $\gt 2$ or $\ge 2$? Finally, having $2$ exponents $\ge 2$ being the same doesn't contradict (at least not directly and in any way I can see) FLT so, as far as I know, that is possible. On the other hand, if it's not possible, it doesn't amount to the FLT. – John Omielan Dec 10 '20 at 01:17
  • From an earlier question I had "For n even, does Fermat Theorem reduce to the Pythagorean Theorem?", the main comment was: "$(a^k,b^k,c^k)$ is a Pythagorean triple, so if you can show that no such pythagorean triple exists for k>1, you would prove Fermat's last theorem for the case of even n>2.". So if at least two exponents (>1) cannot be equal either, is that FLT? (I win the illogical prize!) – vengy Dec 10 '20 at 01:45
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    @vengy FYI, a paper related to what you're asking is arXiv's A note on Pythagorean Triples. With its PDF, for example, Theorem $2.1$ deals with "... the primitive Pythagorean triple generated by $x^m$, $m \in \mathbb{N}$, $m \ge 2$, ...". I have only glanced through this paper but, based on what I've seen, I doubt there will be cases where the same exponent $\gt 1$ occurs more than once, at least with just primitive Pythagorean triples. – John Omielan Dec 10 '20 at 03:05
  • Thanks @John! Good to know and thanks for your insightful ideas too! – vengy Dec 10 '20 at 03:36
  • @vengy You're welcome. FYI, in Pythagorean triples (referred from How does $x^4+y^4=z^2 \implies x^4+y^4=z^4$? ), near the bottom of page $4$, it proves $x^4 + y^4 = z^2$ has no positive integer solutions. This is also asked about & proven in $x^4 + y^4 = z^2$. – John Omielan Dec 10 '20 at 04:23

2 Answers2

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EDIT: The original answer implicitly made the assumption that the sides of the triangle were relatively prime, which is not part of the original post. I've edited to explain why this assumption is valid.

Your question is equivalent to seeking an integer solution to $a^x+b^y=c^z$ in positive integers $a,b,c,x,y,z$, where $x,y,z$ are even and either $x=y\ge 4$ or $y=z\ge 4$. Note that since any $2k$th power is also a square, we can assume WLOG that the third exponent in either case is $2$.

First, let's see that we can reduce to the case where $a,b,c$ are relatively prime. Suppose that our bases have a common divisor $d>1$, and write them as $da,db,dc$ where $a,b,c$ are relatively prime. Then if $x=y$, we have

$$d^x(a^x+b^x)=d^2c^2\implies a^x+b^x=\left(\frac{c}{d^{x/2-1}}\right)^2$$

(Recall that $x$ is even and at least $4$, so $x/2-1$ is a positive integer.) So then $a^x+b^x$, an integer, is equal to the square of the rational number $c/(d^{x/2-1})$. But since the square of a rational number is an integer only when the original rational number is, the tuple $(a,b,c/(d^{x/2-1}))$ is another solution, but this time without any common factors.

A very similar analysis shows that in the case $y=z$, we can reduce to $(a/(d^{z/2-1}),b,c)$ and obtain a solution with relatively prime $a,b,c$. (This sort of reduction does not work if all three exponents are different; if it did, Oscar Lanzi's answer could be reduced to yield a solution with relatively prime $a,b,c$ and falsify Beal's conjecture.)

Having made this reduction, we can find several results in the literature tackling sums of powers of relatively prime positive integers.

The case $x=y$ has been proven impossible by Darmon and Merel (1995); see this citation in Wikipedia.

The case $y=z$ can be reduced to $z=2p$ for some prime $p$, since an $n$th power is an $m$th power if $m$ divides $n$. (Hence, we can ignore $k=8,12,16,18,20,24,28,30,\ldots$.)

$z=4$ follows from a relatively elementary proof by infinite descent, and dates to Fermat (see here for a proof).

$z=6,10$ follow from a result of Bennett, Chen, Dahmen and Yazdani (from Wikipedia's list of partial results on the Beal conjecture.)

As far as I am aware, $z\ge 14$ remains open (and is likely very difficult); one should at least note that the Fermat-Catalan conjecture implies there are at most finitely many solutions, and the ten known solutions to the Fermat-Calatan equation do not have even exponents, so at least no examples are known (very likely none exist).

RavenclawPrefect
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  • Thanks. I've updated the question to include another question about Twin Pythagorean triples. This is my first bounty question, I like your awesome answer, so should I just accept it, and post the other question about Twin triples in another separate question? Don't want to cram too much here. – vengy Dec 12 '20 at 15:33
  • @vengy, yes, I think you should post a separate question. You can of course link to it from here, and backlink from there. – brainjam Dec 12 '20 at 19:22
  • @vengy: Just wanted to bring to your attention that I've edited this answer; the original version referred to results that relied on $a,b,c$ being relatively prime, and I hadn't provided justification for that assumption (which I've now done). – RavenclawPrefect Dec 14 '20 at 02:33
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We do know that all three exponents can assume different values greater than $1$. Here a solution is constructed from the $(3,4,5)$ triple.

Multiply that triple by $9$ to get $(27,36,45)$. This has one leg with exponent $2$ and the other leg with exponent $3$. The hypotenuse has no exponent greater than $1$, but now we take advantage of the fact that multiplying the $(27,36,45)$ triple by any sixth power preserves the exponents of $2$ and $3$ in the legs. We choose $45^6$ as the multiplier and arrive at

$(6075^\color{blue}{3}, 546750^\color{blue}{2},45^\color{blue}{7}).$

Oscar Lanzi
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