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I have found this question answered before but I am stuck going from

$$m^{\phi(n)}+n^{\phi(m)}\equiv1 \pmod{n}\\ n^{\phi(m)}+m^{\phi(n)}\equiv1 \pmod{m}$$ to $$m^{\phi(n)}+n^{\phi(m)}\equiv1 \pmod{mn}$$

Why can we combine $\pmod{m}$ and $(\pmod{n}$ into $\pmod{mn}$? I know it's overall because $\gcd(m,n)=1$, but I don't understand why it is so. Is there a theorem for this specifically? I found this, but this seems to not multiply what is in the modulus together.

Here is the beginning of my proof:

$\gcd(m,n)=1\implies m^{\phi(n)}\equiv1 \pmod{n}$ and $n^{\phi(m)}\equiv1 \pmod{m}$ by Euler's theorem.

$ m^{\phi(n)}\equiv1 \pmod{n} \equiv m^{\phi(n)}+n^{\phi(m)}\equiv1+n^{\phi(m)}\pmod{n}$ and $n^{\phi(m)}\equiv1 \pmod{m}\equiv n^{\phi(m)}+m^{\phi(n)}\equiv1+m^{\phi(n)} \pmod{m}$.

But $m^{\phi(n)} \equiv 0\pmod{m}$ and $n^{\phi(m)}\equiv 0\pmod{n}$.

So we have $m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod{n}$ and $n^{\phi(m)}+m^{\phi(n)}\equiv1 \pmod{m}$...

... which implies $m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod{mn}$.

rtybase
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kaisa
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1 Answers1

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From $$m^{\phi(n)}+n^{\phi(m)}\equiv1 \pmod{n}\\ n^{\phi(m)}+m^{\phi(n)}\equiv1 \pmod{m}$$ we have $$A=m^{\phi(n)}+n^{\phi(m)}-1= n\cdot q_1\\ A=n^{\phi(m)}+m^{\phi(n)}-1=m\cdot q_2$$ Effectively, same number $A$ is divisible by $m$ and $n$, where $\gcd(m,n)=1$. Thus $m\cdot n \mid A$ or $$m^{\phi(n)}+n^{\phi(m)}-1= n\cdot m \cdot q_3 \Rightarrow m^{\phi(n)}+n^{\phi(m)} \equiv 1 \pmod{mn}$$

All this is due to

Proposition. If $\gcd(m,n)=1$, $m \mid A$ and $n \mid A$ then $n\cdot m \mid A$.

Indeed $$A=m\cdot q_1=\color{red}{n\cdot q_2} \tag{1}$$ and from Bezout's lemma $$\exists x,y\in\mathbb{Z}: x\cdot m+y\cdot n=\gcd(m,n)=1$$ Altogether $$x\cdot m\color{blue}{\cdot q_1}+y\cdot n\color{blue}{\cdot q_1}=\color{blue}{q_1} \overset{(1)}{\Rightarrow} \\ x\cdot \color{red}{n\cdot q_2}+y\cdot n\cdot q_1=q_1 \Rightarrow \\ n \mid q_1$$ or $A=m\cdot q_1=m\cdot n\cdot q_3$.

rtybase
  • 16,907