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I'm sorry for the misleading title, my issue is not with the proof for the statement in the title but a step in the proof.

Suppose prime $p$ divides $ab$ but not $a$.

The only positive divisors of $p$ are $1$ and $p$ and $p$ does not divide a gcd$(1,p)=1$.

Therefore $\exists$ integers $r$ and $s :ra+ps=1$.(I do not follow this)

multiplying both sides by $b$ gives $rab+psb=b$

since $p$ divides $rab$ and $psb$, so $p$ divides $ab$.

The proof seems elementary but the statement in bold letters I don't seem to understand the reasoning to it.

abiessu
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Orpheus
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  • Please consider the use of MathJAX and how I have applied it here; feel free to correct any misapplied adjustments as well... – abiessu Aug 09 '21 at 16:41
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    That's called the Bezout identity. The gcd of any two integers $a$ and $b$ can be expressed that way in terms of $a$ and $b$. – Arturo Magidin Aug 09 '21 at 16:41
  • This is a property of the gcd. If $\gcd(a,b)=d$ then there are some $k,l\in\mathbb{Z}$ such that $ka+lb=d$. This property is even more important than the definition of the gcd. (actually, there are even equivalent definitions which involve that property) – Mark Aug 09 '21 at 16:44
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    so you mean the identity was merely applied to a and p and since gcd =1,ra+sp=1? – Orpheus Aug 09 '21 at 16:50

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