两个整数的最大公约数和最小公倍数的乘积等于它们两的乘积

There exists a theorem stating that for any two positive integers $a$ and $b$, the product of their greatest common factor (GCF) and their least common multiple (LCM) is equal to the product $ab$.

Proof of the theorem

We let $a = (2^{a_1}) \cdot (3^{a_2}) \cdot (5^{a_3}) \cdot (\cdots) \cdot (P^{a_n})$ and $b = (2^{b_1}) \cdot (3^{b_2}) \cdot (5^{b_3}) \cdot (\cdots )\cdot (P^{b_n})$, where $P$ is the greatest element of $\{ \text{prime numbers}\}\cap(\{\text{factors of } a\}\cup\{\text{factors of } b\})$, and $n$ is the index of $P$ in $\{\text{primes}\}$.

Lemma 1. The greatest common factor of $a$ and $b$ is $G=(2^{\min(a_1, b_1)}) \cdot (3^{\min(a_2, b_2)}) \cdot (5^{\min(a_3, b_3)}) \cdot (\cdots) * (P^{\min(a_n, b_n)})$. Proof of lemma 1. Obviously, by properties of exponents, $G$ divides both $a$ and $b$. Now, again by properties of exponents, $G$ is the greatest number that divides both $a$ and $b$ (since if you increment one power of one prime number, it will not be a factor of at least one of $a$ or $b$ (for that prime, say $p^*$, let the two powers be $r$ and $s$, where $r \le s$. Then we have that the power of $p^*$ in $p^* G$ is $r+1$, and then $p^* G$ does not divide the number where $r$ is the exponent of $p^*$. Thus we have defined the GCF of two numbers in terms of the numbers' prime factors. Lemma 2. The least common multiple of $a$ and $b$ is $L=(2^{\max(a_1, b_1)}) \cdot (3^{\max(a_2, b_2)}) \cdot (5^{\max(a_3, b_3)}) \cdot (\cdots) * (P^{\max(a_n, b_n)})$. Proof of lemma 2. Obviously, by properties of exponents, $L$ is a multiple of both $a$ and $b$. Now, again by properties of exponents, $L$ is the least(since if you decrement one power of one prime number, it will not be a multiple of at least one of $a$ or $b$ (for that prime, say $p^*$, let the two powers be $r$ and $s$, where $r \ge s$. Then we have that the power of $p^*$ in $G/p^*$ is $r-1$, and then $G/p^*$ is not a multiple of the number where $r$ is the exponent of $p^*$. Thus we have defined the LCM of two numbers in terms of the numbers' prime factors. Lemma 3. We have $\max(x, y) + \min(x, y) = x + y$. Proof of lemma 3. WLOG, assume $x \le y$. Then $x = \min(x,y)$ and $y=\max(x, y)$. The result follows by substitution. The actual proof. We consider $$ab = (2^{a_1 + b_1}) \cdot (3^{a_2 + b_2}) \cdot (5^{a_3 + b_3}) \cdot (\cdots) \cdot (P^{a_n + b_n})$$ and $$GL=(2^{\max(a_1, b_1)+\min(a_1, b_1)}) \cdot (3^{\max(a_2, b_2) + \min(a_2, b_2)}) \cdot (5^{\max(a_3, b_3) + \min(a_3, b_3)}) \cdot (\cdots) * (P^{\max(a_n, b_n) + \max(a_n, b_n)}).$$ By lemma 3, these two are equal. Thus we have $ab =$ GCF$(a,b)$LCM$(a,b).\text{ } \blacksquare$

We've used the gcd/lcm formula with prime factorisation and a fact that $\max \{ x,y \}+ \min \{ x,y \}=x+y$ to prove $\gcd (m,n) \cdot \text{lcm}(m,n)=mn$ (see Andrew Machkasov's proof in other explanation).

Can this be generalised? Is this following generalisation true?

Why then? Because we can't always have $$\max \{ m_1,m_2, \ldots, m_k \}+ \min \{ m_1,m_2, \ldots, m_l \}=m_1+m_2+ \ldots+m_k.$$ In fact, for $k \ge 3$, $LHS<RHS$. Since the case $k=2$ is special, that's why we have the beautiful identity $\gcd (m,n) \cdot \text{lcm}(m,n)=mn$.

Let us first consider the following: $gcd(a, b) = 1 \implies lcm(a, b) = a b$ $lcm(a c, b c) = c lcm(a, b)$ And lastly: $gcd(\frac{a}{gcd(a, b)}, \frac{b}{gcd(a, b)} ) = 1$

Why is it so? GCD indicates how much alike are the numbers (according to their prime factors). Dividing a number $a$ by its GCD with $b$ will result in number $c$ which is nothing like $b$, and so $GCD(c, b) = 1$. So the third identity may be easily understood.

Now the first identity is trivial: if numbers are nothing alike, then LCM must be their product.

The second formula interprets as follows: if there is something alike between numbers $a$ and $b$ then it must surely be a part of their LCM.

Invoking all of them: $lcm(a, b) = lcm(\frac{a gcd(a, b)}{gcd(a, b)}, \frac{b gcd(a, b)}{gcd(a, b)} ) =$ $gcd(a, b) lcm(\frac{a}{gcd(a, b)}, \frac{b}{gcd(a, b)} ) = gcd(a, b) \frac{a}{gcd(a, b)} \frac{b}{gcd(a, b)} = \frac{a b}{gcd(a, b)}$