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.
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2$\begingroup$ For a fixed $c$, your polynomial defines a surface, not a curve, and hence Faltings's theorem does not apply to it. If $c = 3$, it is a rational surface and has infinitely many solutions (most of which have $a \ne b$ and $r \ne s$). If $c = 4$, it is a K3 surface and probably has infinitely many points with $a \ne b$ and $r \ne s$ (a quick search turned up at least $48$). $\endgroup$– Jeremy RouseCommented Jul 20, 2020 at 21:12
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$\begingroup$ @JeremyRouse so you mean it has infinitely many solutions always? $\endgroup$– vidyarthiCommented Jul 20, 2020 at 21:16
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1$\begingroup$ @vidyarthi for $c \geq 5$ the equation you wrote down defines a surface of general type, so it is unlikely to have infinitely many integral points unless they accumulate on some subvariety. $\endgroup$– Stanley Yao XiaoCommented Jul 20, 2020 at 23:28
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3 Answers
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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.
answered Jul 20, 2020 at 23:47
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$\begingroup$ thanks, I think the solution space is not Zariski dense in the surface, is it true? $\endgroup$ Commented Jul 21, 2020 at 10:35
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1$\begingroup$ @vidyarthi it is conjectured to be true by the Bombieri-Lang conjecture, but we are far from being able to prove something like this. The best we can do is that after deleting the "obvious" subvarieties which we expect to contain the bulk of points, the remaining rational points are sparse (in the sense that there are few of them of bounded height) $\endgroup$ Commented Jul 21, 2020 at 11:35
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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.
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$ax^k + by^k = au^k + bv^k\tag{1}$
$a,b,x,y,u,v$ 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: $x^4 + 3y^4 = u^4 + 3v^4$
$(x0,y0,u0,v0)=(4,1,2,3)$.
$(x,y,u,v)=(1068p-1424,-489p+652,-114p+152,837p-1116)$
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$\begingroup$ There's a simpler way to show one solution of $x^4+3y^4=u^4+3v^4$ implies infinitely many; if $(x,y,u,v)$ works, and $k$ is any integer, then $(kx,ky,ku,kv)$ works. $\endgroup$ Commented Sep 27, 2022 at 5:41