The greatest common divisor of a product of two primes

Question

Assume $p,q \in \mathbb{P}$, if $a \in \mathbb{Z}$ and $a \notin \{1,p,q,p\cdot q\}$, then I know that $\gcd(a,p\cdot q)=1$. What I can't seem to do is prove it. Number theory (in my opinion) means having loads of facts at your disposal, most of which I just can't remember (it was a long time since I did this).

Can someone please provide me a proof. My thinking was to try a proof by contradiction, but I couldn't find a contraction.

Thanks in advance

Update: I forgot to add an extra condition to $a$, it must be less than $p\cdot q$ So $a \in \mathbb{Z}$ and $a \notin \{1,p,q,p\cdot q\}$ and $ 0 \leq a \leq p\cdot q$

Answer 1

It can't be proven. With $p=2$, $q=3$ and $a=9$, $gcd(9,6)\neq 1$.

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You must be remembering something else. It is possible to prove that ($gcd(p,a)=gcd(q,a)=1$) iff $gcd(a,pq)=1$.