Why $p_1\mid a \wedge p_2\mid a \Rightarrow p_1p_2\mid a$ where $p_i$ is a prime number, $p_i \neq p_j$ and $a$ is a integer?
I don't fully understand that.
Also, is it true when $p_i$ is something different from a prime number?
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$\begingroup$ This is not true. Take $p_1=p_2=a$. $\endgroup$– Dietrich BurdeNov 10, 2020 at 10:13
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$\begingroup$ I edited the question to say that $p_i \neq p_j$. $\endgroup$– KarolNov 10, 2020 at 10:39
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$\begingroup$ For the second part of the question (non-prime $p_i$), you need still $p_1$ and $p_2$ to be coprime, as an example $p_1=2,p_2=4,a=4$ shows. $\endgroup$– SilNov 10, 2020 at 10:44
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$\begingroup$ @Sil, thank you, below in the answer it's beem already said. $\endgroup$– KarolNov 10, 2020 at 10:51
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Edit
Thanks to Dietrich Burde for indicating that a counter example is the case where $p_1$ and $p_2$ are the same prime. This answer assumes that they are distinct primes.
$p_1 | a \implies ~\exists ~r ~\text{such that} ~a = p_1 \times r.$
Therefore, $p_2 | a ~\implies ~p_2 | (p_1 \times r).$
Since $p_1$ and $p_2$ are relatively prime, this implies that
$p_2 | r \implies ~\exists ~$s$ ~\text{such that} ~r = ~p_2 \times s.$
Therefore, $p_2 \times s \times p_1 = r \times p_1 = a.$
A sufficient (but not necessarily required) condition is that $p_1$ and $p_2$ be relatively prime. They don't have to actually be prime #'s.
answered Nov 10, 2020 at 10:09
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$\begingroup$ But $p_2sp_1=rp_1$ is wrong for $r=s=1$ (the OP doesn't say that $p_1$ and $p_2$ are relatively prime). $\endgroup$ Nov 10, 2020 at 10:14
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$\begingroup$ But you agree that the choice $p_1=p_2=a$ is a counterexample to the statement? So you should need that $p_1$ and $p_2$ are distinct primes. $\endgroup$ Nov 10, 2020 at 10:24
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