How to prove $\gcd(a,\gcd(b, c)) = \gcd(\gcd(a, b), c)$?

Same answer as I just gave in sci.math...

Note that $$d|x,y\Longleftrightarrow d|\gcd(x,y).$$ So: $$\begin{align*} d|a,\gcd(b,c) &\Longleftrightarrow d|a,b,c\\ &\Longleftrightarrow d|\gcd(a,b),c \end{align*}$$

Please note that this solution uses an idea that is very similar to the idea in the solution posted much earlier by ncmathsadist.

We show that for any integer $u$, if $u$ divides the left-hand side, then $u$ divides the right-hand side, and vice-versa. Thus the left-hand side and the right-hand side have the same set of divisors, so must be equal, since they are both non-negative.

Now suppose that $u$ divides $\gcd(a, \gcd(b, c))$. Then $u$ divides $a$ and $u$ divides $\gcd(b,c)$. So $u$ divides $b$ and $c$, and therefore $a$, $b$, and $c$.

Now look at the right-hand side. We know that $u$ divides all of $a$, $b$, and $c$. So $u$ divides $\gcd(a,b)$, and therefore $u$ divides $\gcd(\gcd(a,b),c)$.

Showing that if $u$ divides the right-hand side, then $u$ divides the left-hand side is essentially the same calculation, and can be omitted.