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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 $2n \mid \varphi(m^n+p^n)$

Lord_Farin
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maikhaiok
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    It looks like $p$ can be any natural number not equal to $m$ - it is not necessarily prime? – John Martin Jun 15 '13 at 14:00
  • It suffices to show it for $\gcd(m,p)=1$, because if $\gcd(m,p)=d$ then $\varphi(m^n+p^n)=\varphi(d^n(a^n+b^n))$ with $da=m$ and $db=p$. And since $\gcd(d^n,a^n+b^n)=1$ this equals $\varphi(d^n)\varphi(a^n+b^n)$ and the statement has to be true for $a$ and $b$ too, so the factor $\varphi(d^n)$ is unnecesary. Maybe this helps... – Bart Michels Jun 16 '13 at 13:40
  • @barto It does not follow that $\gcd(d^n,a^n+b^n) = 1$. Take, for example, $m = 2$, $p = 6$; then $d = 2$, $a = 1$, $b = 3$ and you are claiming $2^n$ is coprime to $1^n + 3^n$. – FredH Jun 17 '13 at 05:37

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$\require{begingroup}\begingroup\let\phi\varphi\let\geq\geqslant\let\leq\leqslant\DeclareMathOperator\ord{ord}$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 $\phi(x)\mid\phi(y)$. $\endgroup$

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Bart Michels
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