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There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (https://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.

Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$.

The proof of this uses Dirichlet's theorem on infinitude of primes.

If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.

My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d. $$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial?

The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.

[Edit} I should have added the link https://arxiv.org/abs/math/0409377.

[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.

There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (https://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.

Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$.

The proof of this uses Dirichlet's theorem on infinitude of primes.

If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.

My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d. $$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial?

The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.

[Edit] I should have added the link https://arxiv.org/abs/math/0409377.

[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.

There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (http://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.

Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$.

The proof of this uses Dirichlet's theorem on infinitude of primes.

If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.

My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d. $$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial?

The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.

[Edit} I should have added the link http://arxiv.org/abs/math/0409377.

[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.

There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (https://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.

Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$.

The proof of this uses Dirichlet's theorem on infinitude of primes.

If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.

My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d. $$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial?

The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.

[Edit} I should have added the link https://arxiv.org/abs/math/0409377.

[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.

There is this really nice paper by J.P.Serre on the congruence subgroup property for SL_2 for $S$-arithmetic groups (http://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.

Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient $\phi $ function. for each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers

$$N(a,b)= g.c.d. ( \phi (ax+b): x=1,2,3,\cdots ).$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not: $N(a,b)$ divides $8$.

The proof of this uses Dirichlet's theorem on infinitude of primes. If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.

My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d.

$$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial? The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.

[Edit} I should have added the link http://arxiv.org/abs/math/0409377.

[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.

There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (http://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.

Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$.

The proof of this uses Dirichlet's theorem on infinitude of primes. 

If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.

My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d. $$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial?

The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.

[Edit} I should have added the link http://arxiv.org/abs/math/0409377.

[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.