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When is $\phi(a^n+b^n+c^n)=0\mod n$?

A corollary Zsigmondy's Theorem leads to the following congruence (one can look to $(24)$),$\phi(a^n+b^n)=0\mod n$ whenever $a, b$ are coprime and $n \neq 2$ and $(a,b)\neq(1,1)$. (Here $\phi$ is the ...
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Set of primes $p_{1}\equiv 3 \bmod p_{2}$ such that $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod p_1$ with $p_1,p_2\equiv 3\bmod 4$?

let $p_1$ and $p_2$ be positive primes such that $p_1,p_2 \equiv 3\bmod 4$ and $\phi$ is the Euler totiont function , I want to find the Set of primes $p_{1}\equiv 3 \bmod p_{2}$ such that $\phi(2^...
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$\varphi(m+n)\mid n$ for some positive integer $n$

Let $\varphi$ be Euler's totient function. If $p$ is a prime, then $\varphi(1+n)=n$ for $n=p-1$. Question. Is it true that for each integer $m>1$ there is a positive integer $n\le m^2-m$ such that ...
Zhi-Wei Sun's user avatar
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7 votes
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On $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$

Euler's totient function $\varphi$ is multiplicative, and it plays important roles in number theory. QUESTION: Is it true that for each integer $m>6$ we have $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$...
Zhi-Wei Sun's user avatar
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