Prove that we always have $ 2n \mid \varphi(m^n+p^n) $ For each $ a ∈ \Bbb N^*$, denoted by $\varphi (a) $ is the number of positive integers not exceeding $a$ and coprime to $a$. 
Let $n, m, p ∈ \Bbb N^*, m \ne p$. Prove that we always have $2n \mid \varphi(m^n+p^n)$
 A: $\require{begingroup}\begingroup\let\phi\varphi\let\geq\geqslant\let\leq\leqslant\DeclareMathOperator\ord{ord}$Idea 1: Prove that there exists an element of order $2n$ modulo $m^n+p^n$.
Idea 2: $m$ has order $n$ modulo $m^n-1$.
Indeed, for $x^k$ to be $\equiv1\pmod{m^n-1}$ we need $k\geq n$.
Key idea: $m$ has order $2n$ modulo $m^n+1$.
Indeed, from $m^{2n}\equiv1\pmod{m^n+1}$ we have $\ord_{m^n+1}(m)\mid 2n$. On the other hand, for $m^k$ to be $\equiv1\pmod{m^n+1}$ we need $k>n$. Hence $\ord_{m^n+1}(m)=2n$.
Key idea, generalised: If $\gcd(m,p)=1$ then $mp^{-1}$ (the inverse taken mod $m^n+p^n$) has order $2n$ modulo $m^n+p^n$. (Note that $\gcd(p,m^n+p^n)$ is $1$ so $p^{-1}$ exists.)
Proof.
We have $m^{2n}\equiv p^{2n}$ so $(mp^{-1})^{2n}\equiv1$. On the other hand, $(mp^{-1})^k\equiv1$ implies $m^n+p^n\mid m^k-p^k$, so $k>n$. Hence $\ord_{m^n+p^n}(mp^{-1})=2n$.
Corollary: If $\gcd(m,p)=1$ then $2n\mid\phi(m^n+p^n)$.
Reduction to $\gcd(m,p)=1$:
Let $d=\gcd(m,p)$ and $m=dm_0$, $p=dp_0$ so $\gcd(m_0,p_0)=1$. Then 
$$2n\mid\phi(m_0^n+p_0^n)\mid\phi(d^n\cdot(m_0^n+p_0^n))=\phi(m^n+p^n)$$
because $x\mid y$ implies $\phi(x)\mid\phi(y)$.
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