finding all numbers $a,b$ that divide $c$ but product of $a$ and $b$ does not divide $c$

Question

Let $a,b,c\in \mathbb{Z}$, such that $a$ and $b$ are relatively prime, and both $a$ and $b$ divide $c$. Prove that in this case $ab$ divides $c$. Find all numbers such that $a$ and $b$ divide $c$ but $ab$ does not.

I proved the first part easily. For the second part, is the answer that any two $a,b\in \mathbb{Z}$ which are not relatively prime, their product will not divide $c$?

Answer 1

I don't have the answer but I have an example that disproves your answer to the second part.

Let a = 3, b = 6 and c = 18.

18 = 0 mod 3

18 = 0 mod 6

18 = 0 mod 18

3, 6 are not relatively prime; but their product still divides 18.

Answer 2

For the second part, is the answer that any two $a, b\in \mathbb{Z}$ which are not relatively prime, their product will not divide $c$?

Not exactly. For example, take $a = b = 2$ and $c = 4$.

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However, what you can say is the following:

Let $a, b \in \Bbb Z$ be non-coprime. Then, there exists some $c \in \Bbb Z$ such that $a$ and $b$ both divide $c$ but $ab$ does not.

Note that there exists some $c$. Your earlier phrasing made it sound like it is true for all $c$ that is a multiple of $a$ and $b$ both.

Now, to prove that claim, we simply find a $c$. You can verify that $$c = \dfrac{ab}{\gcd(a, b)}$$ does the job.