Given a prime $\,p\ne3$, is it always possible to find another prime q such that $\,\phi(q)=\phi(p\cdot2^{2n+1})\,$ for some $n\in\mathbb{N}$ ($\,\phi\,$ is the Euler's totient function)?

Some examples are:

$p=2\;\rightarrow\;q=5\;\;\;(n=1)$

$p=7\;\rightarrow\;q=97\;\;\;(n=2)$

$p=31\;\rightarrow\;q=7681\;\;\;(n=4)$

Many thanks.

Given a prime $\,p\ne3$, is it always possible to find another prime q such that $\,\phi(q)=\phi(p\cdot2^{2n+1})\,$ for some $n\in\mathbb{N}$ ($\,\phi\,$ is the Euler's totient function)?

Some examples are:

$p=2\;\rightarrow\;q=5,\,17,\,257,\,65537$

$p=7\;\rightarrow\;q=97,\,6597069766657$

$p=31\;\rightarrow\;q=7681,\,2013265921,\,2061584302081$

Many thanks.

Given a prime $\,p\ne3$, is it always possible to find another prime q such that $\,\phi(q)=\phi(p\cdot2^{2n+1})\,$ for some $n\in\mathbb{N}$ ($\,\phi\,$ is the Euler's totient function)?

Some examples are:

$p=2\;\rightarrow\;q=5$

$p=7\;\rightarrow\;q=97$

$p=31\;\rightarrow\;q=7681$

Many thanks.

Given a prime $\,p\ne3$, is it always possible to find another prime q such that $\,\phi(q)=\phi(p\cdot2^{2n+1})\,$ for some $n\in\mathbb{N}$ ($\,\phi\,$ is the Euler's totient function)?

Some examples are:

$p=2\;\rightarrow\;q=5\;\;\;(n=1)$

$p=7\;\rightarrow\;q=97\;\;\;(n=2)$

$p=31\;\rightarrow\;q=7681\;\;\;(n=4)$

Many thanks.