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.



Is it possible on an elliptic curve both $x,y$ to be arbitrary large powers infinitely often?
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 $(x,y) = (X^{11},Y^7)$.


