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I was considering the Fermat Catalan conjecture, where the equation $a^m+b^n=c^k$ has only finitely many nontrivial solutions (with coprime $a, b, c$) with $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}=1$ (and it is generally believed that the ten known solutions are the only ones). If the sum of the reciprocals if above one, there are families of parametric solutions, and if the sum equals precisely 1, there is only the Catalan solution $2^3+1^6=3^2$

This leads to the obvious (to me) question of what happens when there are four powers? More specifically, my questions are:

  1. Are there always parametric solutions to $a^m\pm b^n\pm c^k=d^l$ if $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}>1$?
  1. If $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}>1$, are solutions generally rare/hard to find?

  2. In the thirteen cases where $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}=1$, are there known solutions?

Certain modular restrictions might force two of the powers to share a prime factor, so I will relax the coprimality constraint to say that the greatest common factor of all four powers is 1.

I am aware of Noam Elkies' solution to $w^4+x^4+y^4=z^4$, which showed that there are infinitely many solutions, and in fact, the solutions are "dense". However, it also seemed to also show that finding solutions to the cases I'm considering is hard, given that the smallest solutions are quite large, and that Euler himself considered the problem and found no solutions. Also, the analogous problem for fifth (and higher) powers is unsolved, leading me to conclude that the sum of reciprocals heuristic is still potentially valid for four powers.

One equation in particular I would like to look into is $w^6+x^6+y^6=z^2$

Edit: It seems like for various values of $(m,n,k,l)$, there are easily derivable solutions which contradict the idea of using the heuristic in full generality. Also, I consider a solution trivial if either one of the powers or a sum of two powers is zero. So I'm going to write out each configuration where the reciprocals sum to one and what is known so far:

$(2,3,7,42):$ Reduces to $x^2+y^3+z^7=1$. Unknown if a parametric solution exists

$(2,3,8,24):$ Reduces to $x^2+y^3+z^8=1$. Unknown if a parametric solution exists

$(2,3,9,18):$ Reduces to $x^2+y^3+z^9=1$. Unknown if a parametric solution exists

$(2,3,10,15):$ Unknown if a parametric solution exists

$(2,3,12,12):$ Reduces to $x^2+y^3+z^{12}=1$. Unknown if a parametric solution exists

$(2,4,5,20):$ Reduces to $x^2+y^4+z^5=1$. Unknown if a parametric solution exists

$(2,4,6,12):$ Reduces to $x^2+y^4+z^6=1$. Unknown if a parametric solution exists

$(2,4,8,8):$ Reduces to $x^2+y^4+z^8=1$. I believe a parametric solution may be derived here from Elkies' work

$(2,5,5,10):$ Reduces to $x^2+y^5+z^5=1$. Unknown if a parametric solution exists

$(2,6,6,6):$ Unknown if any nontrivial solutions exist

$(3,3,6,6):$ Unknown if any nontrivial solutions exist

$(3,4,4,6):$ Parametric solution derived in Max's answer

$(4,4,4,4):$ Elkies has derived infinitely many solutions

It seems like there are two general categories these fall into. The first is equations of the form $x^2+y^3+z^n=1$ or $x^2+y^4+z^n=1$. I feel like these are relatively easy to solve using elliptic curves (although coprimality could be an issue).

The second broad class is the remaining cases of $(2,5,5,10),(2,6,6,6),(3,3,6,6), (3,4,4,6)$, and $(4,4,4,4)$.

I believe that the reason the $(3,4,4,6)$ case was so easy to solve (at least, in the form $(4,3,6,4)$) is that the exponents are coprime. Also, the difference $(a+b)^4-(a-b)^4$ splits very nicely into the two terms $8a^3b$ and $8ab^3$, which aren't difficult to set equal to very high powers of numbers, which Max and Will exploited.

The case I am most interested in now is the $(2,6,6,6)$ case, because it seems to be the most immune to quick tricks involving high common factors.

Update: By considering the elliptic curve $x^3+z^6+1=y^2$, and the trivial solutions $(x,y)=(-z^2,1), (-1,z^3)$, and finding the third solution along the line formed by those two, I have discovered the equation:

$(2z^4+4z^3+5z^2+4z+2)^3+(z^2+z)^6+(z+1)^6=(3z^6+9z^5+15z^4+17z^3+15z^2+9z+3)^2$.

This can provide a solution to the equation $a^6+b^6+c^6=d^2$ if $2z^4+4z^3+5z^2+4z+2$ is a square

Therefore this boils down to the elliptic curve $y^2=2x^4+4x^3+5x^2+4x+2$. This has a rational point (-1,1), however this doesn't correspond to a non-trivial solution of the original equation. Are there other rational points on this curve?

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  • $\begingroup$ What about $a^m+b^n=c^k+d^l$? Is that for some reason less interesting, or perhaps harder? $\endgroup$ Aug 12, 2020 at 14:24
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    $\begingroup$ @Thomas: A. Bremner and M. Ulas proved that equation (2,6,6,6) has infinitely many integer solutions. $\endgroup$
    – Tomita
    Aug 13, 2020 at 2:18
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    $\begingroup$ @Thomas: A. Bremner, M. Ulas, ON $x^{a} ± y^{b} ± z^{c} ± w^{d}= 0, 1/a + 1/b + 1/c + 1/d = 1$ International Journal of Number Theory Vol. 7, No. 8 (2011) 2081-2090 $\endgroup$
    – Tomita
    Aug 13, 2020 at 3:09
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    $\begingroup$ Andrew Bremner, Maciej Ulas, On certain diophantine equations of diagonal type, is freely available at arxiv.org/abs/1311.0717 $\endgroup$ Aug 20, 2020 at 0:45
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    $\begingroup$ Note also discussion at math.stackexchange.com/questions/2767537/… $\endgroup$ Aug 20, 2020 at 0:48

1 Answer 1

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The heurisitc is shady. For example, there are infinitely many coprime solutions for $(m,n,k,l)=(4,6,30,4)$ (where the sum of reciprocals < 1) and for $(m,n,k,l)=(4,6,3,4)$ (where the sum of reciprocals = 1) since $$(3^{90t+30} - 2^{30t+9})^4 + ( 2^{5t+2}\cdot 3^{45t+15} )^6 + (6^{3t+1})^{30} = (3^{90t+30} + 2^{30t+9})^4$$ for any integer $t$.

Perhaps, focusing on pairwise coprime solutions can save from this kind of examples.

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  • $\begingroup$ One can simplify and generalize your identity to $ ( a^{30} - 2^9 b^{30})^4 + ( 2^2 a^{15} b^5)^6 + ( 2 a b^3)^{30} = ( a^{30} + 2^9 b^{30})^4$ for $a$ odd and coprime to $b$. $\endgroup$
    – Will Sawin
    Aug 12, 2020 at 14:19
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    $\begingroup$ Even more generally $(a^{ (n+3) (3n+3) } - 2^n b^{ (n+3) (3n+3) })^4 + ( 2 a^{ 3 (3n+3) } b^{ 3n+3} )^{n+3} + (2 a^{ n+3} b^{ 3(n+3) } )^{3n+3} = (a^{ (n+3) (3n+3) } + 2^n b^{ (n+3) (3n+3) })^4 $ gives quadruples $(4, n+3, 3n+3, 4)$, i.e. with sums of reciprocals arbitrarily close to $\frac{1}{2}$. I guess $a^n + (-a)^n + b^m = b^m$ for $n$ odd gets sum of reciprocals arbitrarily close to $0$. $\endgroup$
    – Will Sawin
    Aug 12, 2020 at 14:22
  • $\begingroup$ @Will: Indeed, there are many variations of this theme. And allowing negative solutions is another Pandora's box. $\endgroup$ Aug 12, 2020 at 15:10
  • $\begingroup$ @MaxAlekseyev your solution is very interesting. I had not considered that there might be simple equations such as that. In regards to pairwise coprimality, that might work in some cases, but modular restrictions come into effect in some cases, like (4,4,4,4), forcing at least two of the powers to be even for example. I don't see how to eliminate the solutions with high common factors without removing all solutions. $\endgroup$
    – Thomas
    Aug 12, 2020 at 21:54
  • $\begingroup$ @WillSawin I quite like your general solution, taking advantage of the difference of two fourth powers. I shall have to think about how those fit in with what I had originally thought. As for $a^n+(-a)^n+b^m=b^m$, I consider that to be trivial, as two of the powers sum to zero. $\endgroup$
    – Thomas
    Aug 12, 2020 at 21:59

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