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This question already has an answer here:

Prove that for all integers $r, s$ and $t$, that $\gcd(\gcd(r, s), t) = \gcd(r, \gcd(s, t))$.

I am stuck in this proof. I have tried using Bézout's Lemma but I have no idea how to proceed further.

Any help would be appreciated.

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marked as duplicate by Arnaud D., GNUSupporter 8964民主女神 地下教會, egreg, Vinyl_cape_jawa, Parcly Taxel Feb 27 at 11:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ And this question and its answers could also be useful. $\endgroup$ – Arnaud D. Feb 27 at 9:57
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You only have to prove that, denoting $\;D(a_1,,\dots,a_n)$ the set of divisors common to $a_1,\dots, n$, one has $$D(\gcd(r,s),t)=D(r,\gcd(s,t))=D(r,s,t).$$

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