Solutions to $(a^c-b^c)+m(r^c-s^c)=0$ in integers Let $c\ge2$  be a fixed positive integer. How many nontrivial solutions in the integers does the equation $(a^c-b^c)+3(r^c-s^c)=0$ have? If $c=2$, I think it has infinitely many solutions as it seems to be similar to Pell's equation. If $c\ge 3$, I think the solutions are finite, by Falting's theorem. Is it that there are no nontrivial solutions to the equation in integers if $c\ge3$ and odd. By nontrivial I mean not all of the integers are zero and neither $a=b$ nor $r=s$.  Any hints?   Thanks beforehand.
 A: Rewriting your equation as, for fixed $m$ and $k$,
$$\displaystyle x^k + my^k = u^k + mv^k, x,y,u,v \in \mathbb{Z},$$
we see that this is of the form $F(x,y) = F(u,v)$ for a binary form of degree $k$ and defines a surface $X_F \subset \mathbb{P}^3$. Heath-Brown showed in this paper that if one deletes the rational lines on this surface, necessarily formed by rational automorphisms of the binary form $F$, then on the remaining open subset $U_1$ of the surface $X_F$ contains at most $O_\epsilon \left(B^{\frac{12k + 16}{9k^2 - 6k + 16} + \epsilon} \right)$ primitive integral points of height at most $B$. This kind of result is the best kind we have in general, as it is not easy to access the geometry of higher degree curves on $X_F$ for arbitrary $F$. The best exponent $\beta_k$ known, which depends only on the degree $k$, is contained in the following paper by myself and Cam Stewart: On the representation of integers by binary forms .
In the special case you are interested in, because the surface is geometrically a Fermat surface which is very special, more can be said about what kind of curves can lie on the surface (which we expect to contribute the bulk of the points). I am not entirely familiar with this but this paper of Browning and Heath-Brown may help.
A: In a joint paper with Andrew Bremner (https://www.sciencedirect.com/science/article/pii/S0022314X13002527#se0070) we obtained several result concerning exitence of rational and integral points on surfaces of the form
$$
a(x^p-y^q)=b(z^r-w^s), 
$$
where $1/p+1/q+1/r+1/s=1$.
If $p=q=r=s=4$ and $h=b/a$, we were able to treat the case of curves that have arithmetic genus 0, hence geometric genus 0, and hence parameterizable, lying on the surface under consideration. If $h$ is not a perfect rational square then there are no curves of arithmetic genus 0.
A: $ax^k + by^k = au^k + bv^k\tag{1}$
$a,b,x,y,u,v$ are integer.
Case $k=3$:
If equation $(1)$ has a known solution, then  equation $(1)$ has infinitely many integer solutions below.
Let $(x0,y0,u0,v0)$ is a known solution.
p,q are arbitrary.
Substitute $x=pt+x0, y=qt+y0, u=pt+u0, v=qt+v0$ to equation $(1)$, then we get
$$t=\frac{-ax0^2p+by0^2q-au0^2p-bv0^2q}{ax0p^2+by0q^2-au0p^2-bv0q^2}$$
Hence  equation $(1)$ has a parametric solution.
Example: $x^3 + 3y^3 = u^3 + 3v^3$
$(x0,y0,u0,v0)=(3,4,6,1)$.
$(x,y,u,v)=(-3(p-q)(2p-3q), -9pq+3q^2+4p^2, -3(p-2q)(p-3q), -9pq+12q^2+p^2)$
p,q are arbitrary.
Without assumption of a known solution , we have another solution below.
$(x,y,u,v)=(3pq-3p^2+3p+8q^2-19q+8, 2q^2+3pq-7q+8-9p+3p^2, -15pq-3p^2+21p-10q^2+35q-28, -25q-15p+20+8q^2+9pq+3p^2)$
p,q are arbitrary.
Case $k=4$:
According to Richmond's theorem for $ax^4 + by^4 + cz^4 + dw^4 = 0$, if equation $(1)$ has a known solution, then  equation (1) has infinitely many integer solutions.
Example: $x^4 + 3y^4 = u^4 + 3v^4$
$(x0,y0,u0,v0)=(4,1,2,3)$.
$(x,y,u,v)=(1068p-1424,-489p+652,-114p+152,837p-1116)$
