Amicable Pair with Smallest Common Prime Factor 5

Theorem
The smallest known amicable pair whose smallest common prime factor is greater than $3$ is the one whose elements are:
 * $m_1 = 5 \times 7^2 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times 101 \times 8643 \times 1 \, 947 \, 938 \, 229$

and:
 * $m_2 = 5 \times 7^2 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times 365 \, 147 \times 47 \, 303 \, 071 \, 129$

This is the smallest counterexample to the observation that:
 * most amicable pair consist of even integers
 * most of the rest, whose elements are odd, have both elements divisible by $3$.

Proof
It is to be demonstrated that these numbers are amicable.

From Sigma of Integer:
 * $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

where:
 * $\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

is the prime decomposition of $n$.

When $k_i = 1$ the individual factor becomes $\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\left({p_i + 1}\right) \left({p_i - 1}\right)} {p_i - 1} = p_i + 1$.

First we make sure we have all the prime factors:

All other factors given are indeed prime.

We establish the contributions to the $\sigma$ values of $m_1$ and $m_2$ by taking the prime factors in turn, and extracting the prime factors of each result.

First, the elements common to both:

This gives a common factor of both $\sigma$ values of:

The remaining prime factors of $m_1$:

Thus:

This gives us the prime decomposition of the rest of $\sigma \left({m_1}\right)$:

The remaining prime factors of $m_2$:

Thus:

This gives us the prime decomposition of the rest of $\sigma \left({m_2}\right)$: