If the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$ can be expressed as $2^m \cdot 3^n$, prove that $m=n$ and find $m$.
This may be considered a sub-problem from another problem. Although I have greatly reduced calculation in my solution, if we can give a simple proof that $m=n$ before we compute them, we'd further cut calculation in my answer in half.
My efforts so far (adapted from my solution to the linked problem):
Case 1: $2^x \equiv 3^y \equiv 1$.
- There are $3$ cases when $3^y\equiv 1$ for each $x \in \{0,3,6,9,12\}$ so the sum of all these $x$'s is $3 \cdot (3+6+9+12)=3 \cdot 30=90$;
- There are $5$ cases when $2^x\equiv 1$ for each $y\in\{0,6,12\}$ so the sum of all these $y$'s is $5\cdot(0+6+12)=90$;
Case 2: $2^x \not \equiv 1$. Note that we must have $3^y \equiv 2,4$ otherwise we can't have $3^y 2^x \equiv 1$.
- There are $2$ cases when $3^y \equiv 2^{-x}$ for each $x\in\{1,2,4,5,7,8,10,11\}$ and the sum of all these $x$'s is $2\cdot(78-30)=96$, where $78=\sum_{i=0}^{12} i$;
- There are $4$ cases when $2^x\equiv 3^{-y}$ for each $y \in \{2,4,8,10\}$ so the sum of all these $y$'s is $4\cdot (2+4+8+10)=96$;
Therefore $m=n=186$.
However I feel that there is a simple argument to just prove $m=n$ so that we can cut our calculation in half. One thing I came up with to prove the sums of $x$'s and $y$'s are equal in Case 1 is as follows:
Denote $o_2, o_3$ as the orders of the elements $2$, $3$, $\pmod 7$, respectively, and $M=12$. Then in Case 1 we need to show $$3\cdot (3+6+9+12) = 5 \cdot (0+6+12)\\ \iff \color{blue}{3} \cdot \color{red}{3} \cdot \color{brown}{\frac{4(4+1)}{2}} = \color{green}{5} \cdot \color{purple}{6} \cdot \color{red}{\frac{2(2+1)}{2}} $$ or equivalently $$\color{blue}{\left(\frac{M}{o_3}+1\right)} \cdot \color{red}{o_2} \cdot \color{brown}{\frac{\frac{M}{o_2} \left(\frac{M}{o_2}+1\right)}{2}} = \color{green}{\left(\frac{M}{o_2} + 1\right)} \cdot \color{purple}{o_3} \cdot \color{red}{\frac{\frac{M}{o_3} \left(\frac{M}{o_3}+1\right)}{2}} \tag 1$$ which is indeed true. And for Case 2 we could do something similar but it would be lengthy and I might as well compute them directly even though there's unnecessary repetition.
My questions:
Is there any better way, e.g., via some simple argument, and better yet combining the two cases together (instead of treating them separately), that $m=n$, $\color{blue}{\it{\text{even before}}}$ we compute them? If it turns out lengthy then it's not worth typing the extra words to replace the few lines of calculation in Case 1 and Case 2 I presented earlier.
Is there a simpler way to solve the original problem by showing $a=b=c$ before computing them?
If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.
