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