Proof
(related to Proposition: Greatest Common Divisors Of Integers and Prime Numbers)
 It follows from the definition of prime numbers that \(p > 1\) and that \(p\) has only the trivial divisors \(1\) and \(p\).
 Therefore the greatest common divisor of \(p\) is \(p\) itself.
 Now, if \(p\) divides the integer \(a\), then the greatest common divisor of \(p\) and \(a\) must be \(p\).
 The first part of the proposition follows: $p\mid a\Rightarrow \gcd(p,a)=p.$
 Now, assume that \(p\not\mid a\).
 Because \(1\) and \(p\) are the only divisors of \(p\), we must have either \(\gcd(p,a)=1\), or \(\gcd(p,a)=p\).
 Because \(p\not\mid a\), only the first is possible.
 This completes the proof
\[p\not a\Longrightarrow \gcd(p,a)=1.\]
∎
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927