The first part is only an introduction how to I find the problem but is not very important for the purpose of find a resolution.
Let $p,q$ be prime number and let $n$ a positive integer number such that $(n,p)=(n,q)=1$. I’m searching $p,q,n$ such that $x^n-1$ has the same number of soluzion in $\mathbb{Q}_p$ and $\mathbb{Q}_q$.
We know, by Hensel lemma, that the polynomial $x^n-1$ has exactly $(p-1,n)$ solution in $\mathbb{Q}_p$ and $(q-1,n)$ solution in $\mathbb{Q}_q$. So I have to solve the equation $$(p-1,n)=(q-1,n)$$ I have no idea how to find a generic solution of this equation, but I found a class of particolar solution: If $p’,q’$ are Sophie Germain primes such that $2p’+1=p$ and $2q’+1=q$ and $n$ is a generic natural number that is not divisible by $p’$ and $q’$ we obtain $$(p-1,n)=(2p’+1-1,n)=(2p’,n)$$ and so this is $1$ if $n$ is odd and this is $2$ if $n$ is even, the same is true for $(q-1,n)$. So a particular solution can be $(7,11,13\cdot 17=221)$ but in this way I have that the number of solution is $1$ or $2$ and it is very big restriction.
But is there a way to find all the solution or a way to find a particular class of solution?