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Notation: $(a,b) = \gcd(a,b)$

If $p,q$ are distinct odd primes, is it true that

$$(pq,(p-1)(q-1)) =1 \iff (pq,\operatorname{lcm}(p-1,q-1))=1\;?$$

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2 Answers 2

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Yes, that statement is true. In fact, it is true for all $p$ and $q$ integers greater than $1$: they do not need to be distinct or odd primes.

The key point is that for positive integers $a$ and $b$, the prime factors of $\operatorname{lcm}(a,b)$ are the same as the prime factors of $ab$, which are also the prime factors of either $a$ or $b$ or both. Let me know if you need a proof of this (there are several).

Therefore, for any positive integers $a,b,c$,

$$\gcd(c,ab)=1 \iff \gcd(c,\operatorname{lcm}(a,b))=1$$

Now substitute $c=pq,\ a=p-1,\ b=q-1$, and we get your proposition.

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Let $(pq,(p-1)(q-1))=1 $, suppose $k=(pq,lcm(p-1,q-1))$ then
$k|pq$ and $k|lcm(p-1,q-1)=\frac{(p-1)(q-1)}{(p-1,q-1)} \Rightarrow$
$k|pq$ and $k|(p-1)(q-1)$ , since $(pq,(p-1)(q-1))=1\Rightarrow$ $k=1$ .

Now let $(pq,lcm(p-1,q-1))=1$ , suppose $(pq,(p-1)(q-1))=k $ then
$k|pq$ and $k|(p-1)(q-1)=pq-p-q+1 \Rightarrow k\nmid (p-1,q-1)$ and $k|(p-1)(q-1)$ then $k|lcm(p-1,q-1)$ since $(pq,lcm(p-1,q-1))=1$ so $k=1$ .

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