GCD and LCM from Prime Decomposition

Theorem
Let $$m, n \in \Z$$.

Let:
 * $$m = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$$;
 * $$n = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}$$;
 * $$p_i \backslash m \lor p_i \backslash n, 1 \le i \le r$$.

That is, the primes given in these decompositions may be divisors of either of the numbers $$m$$ or $$n$$.

Note that if one of the primes $$p_i$$ does not appear in the decompositions of either one of $$m$$ or $$n$$, then its corresponding index $$k_i$$ or $$l_i$$ will be zero.

Then the following results apply:


 * $$\gcd \left\{{m, n}\right\} = p_1^{\min \left\{{k_1, l_1}\right\}} p_2^{\min \left\{{k_2, l_2}\right\}} \ldots p_r^{\min \left\{{k_r, l_r}\right\}}$$


 * $$\operatorname{lcm} \left\{{m, n}\right\} = p_1^{\max \left\{{k_1, l_1}\right\}} p_2^{\max \left\{{k_2, l_2}\right\}} \ldots p_r^{\max \left\{{k_r, l_r}\right\}}$$

Proof
First we prove the result for the GCD:

Let $$d \backslash m$$. Now:


 * $$d$$ is of the form $$p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le h_i \le k_i$$.
 * $$d \backslash n \iff \forall i: 1 \le i \le r, 0 \le h_i \le l_i$$

So $$d \backslash m \land d \backslash n \iff \forall i: 1 \le i \le r, 0 \le h_i \le \min \left\{{k_i, l_i}\right\}$$.

For $$d$$ to be at its greatest, we want the largest possible exponent for each of these primes.

So for each $$i \in \left[{1 \,. \, . \, r}\right]$$, $$h_i$$ needs to equal $$\min \left\{{k_i, l_i}\right\}$$.

Hence the result:

$$\gcd \left\{{m, n}\right\} = p_1^{\min \left\{{k_1, l_1}\right\}} p_2^{\min \left\{{k_2, l_2}\right\}} \ldots p_r^{\min \left\{{k_r, l_r}\right\}}$$


 * We can get the corresponding result for the LCM by using this result:

Sum Less Minimum is Maximum: $$a + b - \min \left\{{a, b}\right\} = \max \left\{{a, b}\right\}$$.

We also make use of Product of GCD and LCM: $$\operatorname{lcm} \left\{{a, b}\right\} \times \gcd \left\{{a, b}\right\} = \left|{a b}\right|$$.

$$ $$ $$ $$

So $$\operatorname{lcm} \left\{{m, n}\right\} = p_1^{\max \left\{{k_1, l_1}\right\}} p_2^{\max \left\{{k_2, l_2}\right\}} \ldots p_r^{\max \left\{{k_r, l_r}\right\}}$$.