Update 31.1.md

docs/Chap31/31.1.md

@@ -121,7 +121,59 @@ $$O(\beta) \times O(\beta) = O(\beta^2).$$
 >
 > $\gcd(a, \gcd(b, c)) = \gcd(\gcd(a, b), c)$.
 
-(Omit!)
+_[The following proof is provided by my friend, Tony Xiao.]_
+
+Let $d = \gcd(a, b, c)$, $a = dp$, $b = dq$ and $c = dr$.
+
+**_Claim_** $\gcd(a, \gcd(b, c)) = d.$
+
+Let $e = \gcd(b, c)$, thus
+
+$$
+\begin{aligned}
+b = es, \\\\
+c = et.
+\end{aligned}
+$$
+
+Since $d \mid b$ and $d \mid c$, thus $d \mid e$.
+
+Let $e = dm$, thus
+
+$$
+\begin{aligned}
+b = (dm)s & = dq, \\\\
+c = (dm)t & = dr.
+\end{aligned}
+$$
+
+Suppose $k = \gcd(p, m)$,
+
+$$
+\begin{aligned}
+            & k \mid p, k \mid m, \\\\
+\Rightarrow & dk \mid dp, dk \mid dm, \\\\
+\Rightarrow & dk \mid dp, dk \mid (dm)s, dk \mid (dm)t, \\\\
+\Rightarrow & dk \mid a, dk \mid b, dk \mid c.
+\end{aligned}
+$$
+
+Since $d = \gcd(a, b, c)$, thus $k = 1$.
+
+$$
+\begin{aligned}
+\gcd(a, \gcd(b, c))
+    & = \gcd(a, e) \\\\
+    & = \gcd(dp, dm) \\\\
+    & = d \cdot \gcd(p, m) \\\\
+    & = d \cdot k \\\\
+    & = d.
+\end{aligned}
+$$
+
+By the claim,
+
+$$\gcd(a, \gcd(b, c)) = d = \gcd(\gcd(a, b), c).$$
 
 ## 31.1-11 $\star$