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Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for an integer $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d\mid n}1$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$ with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).

Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$

Motivated by simple experiments with Pari/GP scripts, I wrote the following claims that seem similar, and that involve other sequences of integers: the known as Mersenne primes and Fermat primes.

Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$

Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$ holds.

Question. I would like to know what is more plausible, if the existence of positive polynomials $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you can to prove, for one of such suitable choices, that the equation $$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.

Many thanks.

Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.

Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for an integer $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d\mid n}1$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$ with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).

Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$

Motivated by simple experiments with Pari/GP scripts, I wrote the following claims that seem similar, and that involve other sequences of integers: the known as Mersenne primes and Fermat primes.

Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$

Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$ holds.

Question. I would like to know what is more plausible, if the existence of different $P(x)\neq Q(x)$ and positive polynomials   $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you can to prove, for one of such suitable choices, that the equation $$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.

Many thanks.

Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.

Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for an integer $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d\mid n}1$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$ with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).

Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$

Motivated by simple experiments with Pari/GP scripts, I wrote the following statement that seem similar, and that involve other sequence of integers: the known as Mersenne primes and Fermat primes.

Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$

Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$ holds for $x=F_n$.

Question. I would like to know what is more plausible, if the existence of positive polynomials $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you one can to prove, for one of such suitable choice, that the equation $$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.

Many thanks.

Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.

Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for an integer $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d\mid n}1$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$ with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).

Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$

Motivated by simple experiments with Pari/GP scripts, I wrote the following claims that seem similar, and that involve other sequences of integers: the known as Mersenne primes and Fermat primes.

Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$

Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$ holds.

Question. I would like to know what is more plausible, if the existence of positive polynomials $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you can to prove, for one of such suitable choices, that the equation $$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.

Many thanks.

Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.

Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for integers $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d|\mid n}d$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$ with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).

Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$

Motivated by simple experiments with Pari/GP scripts, I wrote the following statement that seem similar, and that involve other sequence of integers: the known as Mersenne primes and Fermat primes.

Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$

Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$ holds for $x=F_n$.

Question. I would like to know what is more plausible, if the existence of positive polynomials $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you one can to prove, for one of such suitable choice, that the equation $$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.

Many thanks.

Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.

Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for an integer $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d\mid n}1$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$ with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).

Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$

Motivated by simple experiments with Pari/GP scripts, I wrote the following statement that seem similar, and that involve other sequence of integers: the known as Mersenne primes and Fermat primes.

Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$

Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$ holds for $x=F_n$.

Question. I would like to know what is more plausible, if the existence of positive polynomials $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you one can to prove, for one of such suitable choice, that the equation $$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.

Many thanks.

Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.