On the Diophantine equation $x^{4}+y^{4}=z^p$

Question

Do there exist integers $x,y,z$ with $xyz\neq 0$, such that

$$x^4 + y^4 = z^p$$

where $p\geq 5$ is some prime ?

If yes, are there infinitely many of them ? And if there exists infinitely many of them, what are their parametrizations ?

Answer 1

There are no primitive solutions, even when one of the fourth powers is replaced by a square. This was proved by Bennett, Ellenberg, Ng (see Int. J. Number Theory 6 (2010), 311-338).

Non-primitive solutions there are plenty, see Noam Elkies's comment for an example.

P.S. I found the paper online here, I don't know how long the link lasts.

Answer 2

Our recent paper "Diophantine equations with three monomials" https://doi.org/10.1016/j.jnt.2023.06.011 gives a method for parametrizing all integer solutions to a large class of three-monomial equations, that is, equations of the form $$ a \prod_{i=1}^n x_i^{\alpha_i} + b \prod_{i=1}^n x_i^{\beta_i} = c \prod_{i=1}^n x_i^{\gamma_i}, $$ where $x_1,\dots,x_n$ are variables, $\alpha_i$, $\beta_i$ and $\gamma_i$ are non-negative integers, and $a,b,c$ are non-zero integer coefficients. We call a solution $x=(x_1,\dots,x_n)$ trivial if $\prod_{i=1}^n x_i = 0$ and non-trivial otherwise. The trivial solutions are easy to describe, so we may concentrate on finding the non-trivial ones.

A simple sufficient (but not necessary!) condition that guarantees that our method works is that the system of equations $$ \sum_{i=1}^n \alpha_i z_i = \sum_{i=1}^n \beta_i z_i = \sum_{i=1}^n \gamma_i z_i - 1, \quad \quad z_i \geq 0, \quad i=1,\dots, n, $$ has solutions in non-negative integers $z_1,\dots, z_n$. The answer is particularly simple if the system $$ \sum_{i=1}^n \alpha_i t_i = \sum_{i=1}^n \beta_i t_i = \sum_{i=1}^n \gamma_i t_i + 1, \quad \quad t_i \geq 0, \quad i=1,\dots, n, $$ is also solvable in non-negative integers $t_1, \dots, t_n$. In this case all non-trivial integer solutions to the original equation are given by $$ x_i = \left(a \prod_{j=1}^n u_j^{\alpha_j} + b \prod_{j=1}^n u_j^{\beta_j}\right)^{z_i} \left(c \prod_{j=1}^n u_j^{\gamma_j}\right)^{t_i} w^{-z_i-t_i} \cdot u_i, \quad i=1,\dots, n, $$ where $u_i$ are arbitrary integers and $w$ is any common divisor of the expressions in the brackets.

As a very special case, we can describe all integer solutions to the generalized Fermat equation $$ a x^q + b y^r = c z^p, $$ provided that at least one of the exponents $p,q,r$ is co-prime with the other two. In your case, $(q,r,p)=(4,4,p)$, exponent $p$ is co-prime with the other two, hence the method works.

Let us apply it with, for example, $p=5$. We have $n=3$ variables $(x,y,z)=(x_1,x_2,x_3)$, coefficients $a=b=c=1$, and exponents $\alpha_1=\beta_2=4$, $\gamma_3=5$, all other zeros. The systems for $z_i$ and $t_i$ in this case are $$ 4z_1 = 4z_2 = 5z_3-1 $$ and $$ 4t_1 = 4t_2 = 5t_3+1 $$ These system have solutions in non-negative integers, the simplest ones are $(z_1,z_2,z_3)=(1,1,1)$ and $(t_1,t_2,t_3)=(4,4,3)$. Hence, $$ x = \frac{(u_1^4 + u_2^4) u_3^{20}}{w^{5}} u_1, \quad y = \frac{(u_1^4 + u_2^4) u_3^{20}}{w^{5}} u_2, \quad z= \frac{(u_1^4 + u_2^4) u_3^{16}}{w^{4}}, $$ where $u_!,u_2,u_3$ are arbitrary integers, and $w$ is any common divisor of $u_1^4+u_2^4$ and $u_3^5$. The final answer can be simplified by introducing new parameter $u=u_3^4$.