For positive integers \(a,b,c\), prove that \[ \gcd\bigl(a,\operatorname{lcm}(b,c)\bigr) = \operatorname{lcm}\bigl(\gcd(a,b),\gcd(a,c)\bigr). \]
Solution
Fix a prime \(p\) and write \[ A=v_p(a), \qquad B=v_p(b), \qquad C=v_p(c). \]
1. One has\[ v_p\!\left( \gcd\bigl(a,\operatorname{lcm}(b,c)\bigr) \right) = \min\bigl(A,\max(B,C)\bigr). \]
For positive integers, \[ v_p(\gcd(r,s)) = \min(v_p(r),v_p(s)) \] and \[ v_p(\operatorname{lcm}(r,s)) = \max(v_p(r),v_p(s)). \] Apply these formulas first to \(\operatorname{lcm}(b,c)\) and then to the outer gcd.
2. One has\[ v_p\!\left( \operatorname{lcm}\bigl(\gcd(a,b),\gcd(a,c)\bigr) \right) = \max\bigl(\min(A,B),\min(A,C)\bigr). \]
Apply the same two valuation formulas, first to the two gcds and then to the outer lcm.
3. For all real numbers \(A,B,C\),\[ \min\bigl(A,\max(B,C)\bigr) = \max\bigl(\min(A,B),\min(A,C)\bigr). \]
Interchange \(B\) and \(C\) if necessary and assume \[ B\leq C. \] Then \[ \max(B,C)=C. \] Also \[ \min(A,B)\leq\min(A,C), \] so \[ \max\bigl(\min(A,B),\min(A,C)\bigr) = \min(A,C). \] This equals \[ \min\bigl(A,\max(B,C)\bigr). \]
4. The two integers in the statement have the same \(p\)-adic valuationfor every prime \(p\).
Combine steps <1>1–<1>3.
5. Therefore\[ \boxed{ \gcd\bigl(a,\operatorname{lcm}(b,c)\bigr) = \operatorname{lcm}\bigl(\gcd(a,b),\gcd(a,c)\bigr) }. \]
Two positive integers are equal exactly when their \(p\)-adic valuations agree for every prime \(p\). Step 4 gives that agreement.
6. Q.E.D.
Step 5 is the required identity.