Return to Answer

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

If you want $f$ irreducible, take $d=3$ and $f$ with real irrational root, then use Voloch's answer here

To the edited question about uniform bound.

I think again no.

In Siksek's answer replace $x-2y$ by $x-2^{2^k}y$. You have the solution $(2^{2^k},1)$ and all the other solutions.

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

If you want $f$ irreducible, take $d=3$ and $f$ with real irrational root, then use Voloch's answer here

To the edited question about uniform bound.

I think again no.

In Siksek's answer replace $x-my$ by $x-m^{2^k}y$. You have the solutions $(m^{2^k},1)$ for fixed $h$.

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

If you want $f$ irreducible, take $d=3$ and $f$ with real irrational root, then use Voloch's answer here

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

If you want $f$ irreducible, take $d=3$ and $f$ with real irrational root, then use Voloch's answer here

To the edited question about uniform bound.

I think again no.

In Siksek's answer replace $x-2y$ by $x-2^{2^k}y$. You have the solution $(2^{2^k},1)$ and all the other solutions.

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

If you want $f$ irreducible, take $d=3$ and $f$ with real irrational root, then use Voloch's answer here