Theorem: Amicable Pair with Smallest Common Prime Factor 5

An amicable pair whose smallest common prime factor is greater than $3$ has the elements:
:$m_1 = 5 \times 7^2 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times 101 \times 8693 \times 19 \, 479 \, 382 \, 229$
and:
:$m_2 = 5 \times 7^2 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times 365 \, 147 \times 47 \, 307 \, 071 \, 129$
This was at one point the smallest known counterexample to the observation that:
:most amicable pairs 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 Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
where:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} =  p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
is the prime decomposition of $n$.
When $k_i = 1$ the individual factor becomes $\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\paren {p_i + 1} \paren {p_i - 1} } {p_i - 1} = p_i + 1$.
We establish the contributions to the divisor sum 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:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} 5
      | r = 6
      | c = Divisor Sum of Prime Number
}}
{{eqn | r = 2 \times 3
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {7^2}
      | r = \dfrac {7^3 - 1} {7 - 1}
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = \dfrac {343 - 1} 6
      | c = 
}}
{{eqn | r = 57
      | c = 
}}
{{eqn | r = 3 \times 19
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {11^2}
      | r = \dfrac {11^3 - 1} {11 - 1}
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = \dfrac {1331 - 1} {10}
      | c = 
}}
{{eqn | r = 133
      | c = 
}}
{{eqn | r = 7 \times 19
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {13}
      | r = 14
      | c = Divisor Sum of Prime Number
}}
{{eqn | r = 2 \times 7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {17}
      | r = 18
      | c = Divisor Sum of Prime Number
}}
{{eqn | r = 2 \times 3^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {19^3}
      | r = \dfrac {19^4 - 1} {19 - 1}
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = \dfrac {130 \, 321 - 1} {18}
      | c = 
}}
{{eqn | r = 7240
      | c = 
}}
{{eqn | r = 2^3 \times 5 \times 181
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {23}
      | r = 24
      | c = Divisor Sum of Prime Number
}}
{{eqn | r = 2^3 \times 3
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {37}
      | r = 38
      | c = Divisor Sum of Prime Number
}}
{{eqn | r = 2 \times 19
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {181}
      | r = 182
      | c = Divisor Sum of Prime Number
}}
{{eqn | r = 2 \times 7 \times 13
      | c = 
}}
{{end-eqn}}
This gives a common factor of both $\sigma_1$ values of:
{{begin-eqn}}
{{eqn | r = 6 \times 57 \times 133 \times 14 \times 18 \times 7240 \times 24 \times 38 \times 182
      | o = 
}}
{{eqn | r = 2^{11} \times 3^5 \times 5 \times 7^3 \times 13 \times 19^3 \times 181
      | c = 
}}
{{end-eqn}}
The remaining prime factors of $m_1$:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {101 \times 8693 \times 19 \, 479 \, 382 \, 229}
      | r = \paren {101 + 1} \paren {8693 + 1} \paren {19 \, 479 \, 382 \, 229 + 1}
      | c = Divisor Sum of Square-Free Integer
}}
{{eqn | r = \paren {2 \times 3 \times 17} \paren {2 \times 3^3 \times 7 \times 23} \paren {2 \times 3 \times 5 \times 7 \times 11^2 \times 37 \times 20 \, 719}
      | c = 
}}
{{eqn | r = 2^3 \times 3^5 \times 5 \times 7^2 \times 11^2 \times 17 \times 23 \times 37 \times 20 \, 719
      | c = 
}}
{{end-eqn}}
The remaining prime factors of $m_2$:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {365 \, 147 \times 47 \, 307 \, 071 \, 129}
      | r = \paren {365 \, 147 + 1} \paren {47 \, 307 \, 071 \, 129 + 1}
      | c = Divisor Sum of Square-Free Integer
}}
{{eqn | r = \paren {2^2 \times 3^4 \times 7^2 \times 23} \paren {2 \times 3 \times 5 \times 11^2 \times 17 \times 37 \times 20 \, 719}
      | c = 
}}
{{eqn | r = 2^3 \times 3^5 \times 5 \times 7^2 \times 11^2 \times 17 \times 23 \times 37 \times 20 \, 719
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {m_1} = \map {\sigma_1} {m_2}
      | r = \paren {2^{11} \times 3^5 \times 5 \times 7^3 \times 13 \times 19^3 \times 181} \paren {2^3 \times 3^5 \times 5 \times 7^2 \times 11^2 \times 17 \times 23 \times 37 \times 20 \, 719}
      | c = 
}}
{{eqn | r = 2^{14} \times 3^{10} \times 5^2 \times 7^5 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times 20 \, 719
      | c = 
}}
{{end-eqn}}
It remains to be shown that:
:$\map {\sigma_1} {m_1} = \map {\sigma_1} {m_2} = m_1 + m_2$
So:
{{begin-eqn}}
{{eqn | l = m_1 + m_2
      | r = 5 \times 7^2 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times \paren {101 \times 8693 \times 19 \, 479 \, 382 \, 229 + 365 \, 147 \times 47 \, 307 \, 071 \, 129}
      | c = 
}}
{{eqn | r = 6921698310331805 \times \paren {17102761241386397 + 17274035101540963}
      | c = 
}}
{{eqn | r = 6921698310331805 \times 34376796342927360
      | c = 
}}
{{eqn | r = 237945813161460881075326612684800
      | c = 
}}
{{eqn | r = 2^{14} \times 3^{10} \times 5^2 \times 7^5 \times 11^2 \times 13 \times 17 \times 19^3 \times 23 \times 37 \times 181 \times 20719
      | c = 
}}
{{end-eqn}}
{{qed}}