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Let $f(x,y)=0$ be irreducible elliptic curve over the rationals.

Are there $f$ for which:

Both $x,y$ are arbitrary large powers infinitely often, i.e. infinitely many rational points $(u^k,v^m)$ with both $k,m$ arbitrary large?

For $x,y$ squares (asked in previous revision) this is possible. Take $f(x,y)=x^{6} - 2 x^{3} y^{3} + y^{6} - 72 x^{3} - 72 y^{3} + 1296$.

$f(x^2,y^2)$ is divisible by $x^3 + y^3 - 6$ which is genus $1$ of positive rank.

For large $k,m$ the n-conjecture implies the number of monomials can't be too small.

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  • $\begingroup$ Squares of what? $x$ is a rational function on the curve with divisors of zeroes (or poles) of degree 3, it cannot be a square of a rational function. $\endgroup$
    – abx
    Commented Jan 3, 2015 at 13:04
  • $\begingroup$ @abx I mean squares of rationals $x=u^2,y=v^2$. I don't ask about rational function at all. $\endgroup$
    – joro
    Commented Jan 3, 2015 at 13:06
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    $\begingroup$ You are essentially looking for rational points on a curve $f(x^2,y)=0$ or $f(x^2,y^2)=0$ depending on your condition. So you have to work out if the curve has genus 1 or > 1, which depends on whether the obvious cover is unramified or not. This is not an MO question. $\endgroup$
    – Felipe Voloch
    Commented Jan 3, 2015 at 13:15
  • $\begingroup$ @FelipeVoloch If you say so. The problem is I don't know $f$ and there are many of them. $\endgroup$
    – joro
    Commented Jan 3, 2015 at 13:20
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    $\begingroup$ Usually the cover will be ramified, so there will be only finitely many solutions. For example, if you have a Weierstrass model $y^2 = x^3 + ax + b$, then $y^2 = x^6 + ax^2 + b$ defines a curve of genus 2 unless $b = 0$. $\endgroup$
    – Michael Stoll
    Commented Jan 3, 2015 at 13:29

2 Answers 2

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The method joro suggests works with little change for any $(k,m)$.

Fix an elliptic curve $E: P(X,Y)=1$ of positive rank. Some simple examples are $F(X,Y) = Y^2 - X^3 + 2X$ and $F(X,Y) = Y^2 - X^3 - 2$, each with generator $(X,Y)=(-1,1)$.

Let $u$ and $v$ be "random" rational functions on $E$ that generate the function field (almost any choice will work, see below). Then $u$ and $v$ satisfy a minimal equation $g(u,v)=0$ which defines an algebraic curve birational to $E$ and thus has infinitely many rational solutions.

Set $x = u^k$ and $y = v^m$. As long as these functions, too, generate the function field of $E$, the equation $f(x,y)=0$ that they satisfy gives an irreducible elliptic curve with infinitely many rational points at which $x$ is a $k$-th power and $y$ is an $m$-th power. To check this it is enough to verify that the map $(X,Y) \mapsto (x,y) = (u^k,v^m)$ is generically injective on $E$, and for that it's enough to find one point $(u,v)$ on $E$ that's not a zero or pole of either $u$ or $v$ and such that there's no other $(u',v') \in E$ with ${u'}^k=u^k$ and ${v'}^m = v^m$. For some $k$ and $m$ we can even take $(u,v) = (X,Y)$, and in general a pair of "random" translates $(X+u_0, Y+v_0)$ should suffice.

The polynomial equation $f(x,y) = 0$ can be computed by eliminating $u,v$ from the system $g(u,v) = u^k-x = v^m-y = 0$: take the resultant of the first two equations with respect to $u$ to obtain a relation between $v$ and $x$, and then take the resultant w.r.t. $v$ of that relation and $v^m-y$.

For example, applying this recipe to $Y^2 = X^3 + 2$ with $(k,m)=(11,7)$ and $(u,v)=(X,Y)$ can be done with the gp command

polresultant(polresultant(Y^2-X^3-2, X^11-x, X), Y^7-y, Y)

which yields the irreducible polynomial

x^21 + (-36960*y^2 + 14336)*x^18 + (154*y^6 + 396506880*y^4 + 19029491712*y^2 + 88080384)*x^15 + (12029248*y^8 - 1216366964736*y^6 + 284451691560960*y^4 - 391801278038016*y^2 + 300647710720)*x^12 + (5236*y^12 + 41292382208*y^10 + 649374955536384*y^8 + 261188560760078336*y^6 + 4032263773430480896*y^4 + 1338427807910330368*y^2 + 615726511554560)*x^9 + (-61949888*y^14 + 5483890384896*y^12 - 20842181162958848*y^10 + 8440505090430730240*y^8 - 456851090435200778240*y^6 + 2810028717064564244480*y^4 - 898304588698945060864*y^2 + 756604737398243328)*x^6 + (16016*y^18 + 1301780480*y^16 + 5890099511296*y^14 + 5689217111818240*y^12 + 1688064484691673088*y^10 + 172250320914033410048*y^8 + 5873494991874845310976*y^6 + 55086593993714775883776*y^4 + 78432673497651496353792*y^2 + 516508834063867445248)*x^3 + (-y^22 + 1408*y^20 - 901120*y^18 + 346030080*y^16 - 88583700480*y^14 + 15874199126016*y^12 - 2031897488130048*y^10 + 185773484629032960*y^8 - 11889503016258109440*y^6 + 507285462027012669440*y^4 - 12986507827891524337664*y^2 + 151115727451828646838272)

vanishing on infinitely many pairs $(x,y) = (X^{11},Y^7)$.

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  • $\begingroup$ Is typo in the resultants possible? You don't appear to use $F$, might be wrong. $\endgroup$
    – joro
    Commented Jan 10, 2015 at 7:06
  • $\begingroup$ You're welcome. I think the resultant is right. I just edited to add the gp code. I also checked that a few multiples of $(-1,1)$ yield rational multiples whose coordinates are the expected $11$th and $7$th powers. $\endgroup$
    – Noam D. Elkies
    Commented Jan 10, 2015 at 16:53
  • $\begingroup$ This indeed works. Better proof than checking is to factor q(x^11,y^7) and see it is divisible by the curve. Is it coincidence that in the factorization the degree 18 factor is genus 1 too? $\endgroup$
    – joro
    Commented Jan 11, 2015 at 7:08
  • $\begingroup$ By the way, for $k=m=2$ the resultant is square and is genus 0. $\endgroup$
    – joro
    Commented Jan 11, 2015 at 8:49
  • $\begingroup$ That's covered by the $(u',v')$ test I gave. When $m$ is even you can't take $u,v$ to be the $X,Y$ of the classical Weierstrass equation because then the map $(u,v) \mapsto (u^k,v^m)$ is not generically $1:1$ (try $(u',v') = (u,-v)$). But $(X,Y+1)$ works. (Also for $Y^2=X^3+2$ if $3 \mid k$ you must tweak $u$ to avoid triplication on the $X$ side.) $\endgroup$
    – Noam D. Elkies
    Commented Jan 11, 2015 at 15:30
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Partial answer, someone possibly might hopefully complete it.

Solutions exists for $k=m \in \{2,3,4,5\}$.

The case $k=4$ was found with rational surface and $5$ using very similar approach.

Fix $k,m$ large.

Consider the rational surface given by parametrization: $x=u^k, y=v^m, z=h(u,v)$ where $h(u,v)=0$ is genus $1$, positive rank (possibly better cubic model).

The surface must satisfy some polynomial $g(x,y,z)=0$.

Take $u,v$ on $z=h(u,v)=0$ and $f(x,y)=g(x,y,0)$.

If $f(x,y)=0$ is irreducible, all $x,y$ are large perfect powers and the genus is zero or one.

Experimentally genus zero is possible, though haven't seen it on $h$ cubic model.

So to complete the proof, $f(x,y)=0$ must be irreducible and genus one.

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