The question asks what is known about integer solutions $(\mathbf{x}, \mathbf{y}) = ((x\_1, x\_2, x\_3), (y\_1, y\_2, y\_3))$ to the equation


$$\displaystyle x\_1^{r-1} y\_1^r + x\_2^{r-1} y\_2^r = x\_3^{r-1} y\_3^r$$


where $r \geq 4$ is an integer, $x\_1 x\_2 x\_3 y\_1 y\_2 y\_3 \ne 0$, and $\gcd(x\_1 y\_1, x\_2 y\_2, x\_3 y\_3) = 1$. In particular, does one expect infinitely many solutions, and can one prove an asymptotic lower bound?


When $r = 2,3$ this problem has been studied by [Le Boudec](https://doi.org/10.1093/imrn/rnu255) and [Browning/Valckenborgh](https://projecteuclid.org/download/pdf_1/euclid.em/1338430831) respectively, which answers the question in the affirmative in these cases. Le Boudec establishes the exact asymptotic order of magnitude, while Browning-Valckenborgh establishes a lower bound which is expected to be the correct order of magnitude and an upper bound of $O\_\epsilon(T^{3/5 + \epsilon})$.


As Henri Cohen remarks below, when $r = 4$ it is possible to choose $y\_1 = a, y\_2 = b, y\_3 = c$ in such a way that the genus one plane cubic curve $a^4 x\_1^3 + b^4 x\_2^3 = c^4 x\_3^3$ has a rational point and has positive rank as an elliptic curve (in particular, $y\_1 = y\_2 = 1, y\_3 = 9$ gives one such choice). This trick will not work when $r$ is sufficiently large.



Coprime integer solutions to $x_1^{r-1} y_1^r + x_2^{r-1} y_2^r = x_3^{r-1} y_3^r$
Assuming the abc conjecture, there are only finitely many solutions with $r\geq 5$. Indeed, more generally, abc conjecture implies there are only finitely many sums of the form $a+b=c$ in which $a,b,c$ are coprime and all prime exponents in them are at least $4$. Indeed, for any such triple we have
$$rad(abc)\leq (abc)^{1/4}\leq c^{3/4}$$
(assuming $a,b,c$ are positive, if not you can switch signs and permute them.) Therefore one definitely shouldn't expect there to be infinitely many solutions for $r\geq 5$. However I doubt an unconditional proof is known, even for a fixed value of $r$.


