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I am looking at the following question from my undergraduate Number Theory textbook:

Show that if p,q are different odd primes, and if gcd(p,q)=1, then a$\Phi$(pq)/2 $\equiv$ $1$ mod $pq$.

So far, the approach I have taken is trying to split up a$\Phi$(pq)/2 = (a$\Phi$(p)/2)$\Phi$(q) and apply Euler's theorem but I don't think I am really getting anywhere.

Any help would be greatly appreciated, thank you!

Liv
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3 Answers3

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Note: If $\gcd(a,pq) \neq 1$, then clearly $a^{ \phi (pq) / 2 } \neq 1 \pmod{pq}$. So, I'm assuming that $\gcd(a,pq) = 1$.

Hint: $\phi (pq) = (p-1)(q-1)$, where both of the terms on the right are even, since they are odd primes.

Hence $a^{\frac{(p-1)(q-1) } {2} } \equiv \left( a^{p-1} \right) ^ { \frac{q-1}{2} } \equiv 1 \pmod{p}$.
Similarly for $\pmod{q}$, and hence for $\pmod{pq}$ (since they are distinct primes).

Calvin Lin
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  • @CalvinLin thank you! I think I thought about this problem for so long that I got lost in it, haha. The hint was very helpful to get me back on track. – Liv Nov 13 '19 at 04:09
  • @Liv You might find it helpful to view this in terms of basic $\rm lcm$ laws - which then leads the way to obvious generalizations that often prove handy, e.g. see my answer. – Bill Dubuque Nov 13 '19 at 05:13
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Hint $\ $ It is the special case: $\ \ \begin{align} &\ \ \ m,\ M,\ \ \, n,\,\ N,\ d\\ =\ &\phi(p),\,p,\,\phi(q),\,q,\ \ 2\end{align}\ \,$ in the $\rm lcm$-based generalization below

$\!\begin{align}\text{The proof below shows that:} \ \ \ \ \color{#c00}{a^{\large m}}&\equiv \color{#c00}1\pmod{M}\\ a^{\large n} &\equiv 1\pmod {N}\end{align}\, \Rightarrow\ a^{\large\color{#0a0}{{\rm lcm}(m,n)}}\!\equiv 1\pmod{{\rm lcm}(M,N})$

${\rm so}\ \ d\mid m,n\,\Rightarrow\,m,n\mid mn/d\, \Rightarrow\ \color{#0a0}{{\rm lcm}(m,n)}\mid mn/d\ \Rightarrow\, \bbox[5px,border:1px solid #c00]{a^{\large mn/d}\ \equiv\ 1\ \ \pmod{{\rm lcm}(M,N})}$

by applying $\color{#90f}{\rm MOR}$ = Modular Order Reduction $ $ [or directly: $\,a^{\large {\color{#0a0}{\ell}}}\equiv 1\,\Rightarrow\, a^{\large \ell\:\! k}\!\equiv (a^{\large\color{#0a0}{\ell}})^{\large k}\!\equiv 1^{\large k}\equiv 1$]

Proof $\ $ Let $\,\ell={\rm lcm}(m,n).\,$ Then $\ m\mid\ell\ $ so $\ \color{#c00}{a^{m} \equiv 1}\,\Rightarrow a^{\ell}\equiv 1\pmod{\!M},\,$ again by $\color{#90f}{\rm MOR}$.

Same $\!\bmod N,\,$ so $\, M,N\mid a^{\ell}-1\,\Rightarrow\,{\rm lcm}(M,N)\mid a^{\ell} -1\ $ [or we can use CCRT $ $ vs. $\,\rm lcm\!$ ]

Bill Dubuque
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Hint: $x\cong1\pmod p$ and $x\cong1\pmod q$ implies $x\cong1\pmod{pq}$. For we have $kp+1=x=lq+1\implies kp=lq\implies q\mid kp\implies q\mid k$, by Euclid's lemma.