I have pairwise relatively prime positive integers $a$, $b$, $c$, and $d$ such that $$ \frac{a-b}{4} = \frac{2a-9c}{7} = 27d-10a = 9c-2b = \frac{27d-10b}{41} = \frac{3d-5c}{4} \tag{$\star$} $$ and $$ \frac{a+b}{3} = \frac{2(a+6c)}{7} = 2(7a-18d) = 2(6c-b) = \frac{2(7b+18d)}{41} = \frac{7c+3d}{4}. \tag{$\star\star$} $$
I know, a priori, that the problem I’m working on has exactly one solution $(a,b,c,d)=(29,1,1,11)$.
QUESTION #1: Do ($\star$) and ($\star\star$), independently or together, provide enough information to find the exact numeric solution? Or even just prove $b=c$?
QUESTION #2: If I can also provide, for each pair of variables, an equation of the form $pa^2+qab+rb^2+s=0$, where $p,q,r,s$ are integer constants, would that be enough information to find the exact solution?
I’ve tried everything I know how to throw at it, and just get caught going around in circles.
EDIT #1: Doing a brute-force computer search of pairwise relatively prime odd integers $a,b,c,d$ with $1 \le a \le 1001$ and $1 \le d \le \lceil \tfrac{7}{18}a \rceil$ and $1 \le c \le \lceil \tfrac{2}{9}a \rceil$ and $1 \le b \le a-2$ reveals a number of possible solutions… but adding in the one extra condition $3bd-ac=4$ reduces the set to the desired single solution.
EDIT #2: Expanding the search, there is a solution whenever $a$ is a Pell number $P_{12k-7}$.
$\frac{a-b}{4} = \frac{2a-9c}{7} = 27d-10a = 9c-2b = \frac{27d-10b}{41} = \frac{3d-5c}{4} \tag{$(1)$}$
Above equation (1) has another numerical solution:
$(a,b,c,d)=(w,5w,w,3w)$
Where, $w=41$
Also above has, $(a=c)$
Regarding the second simultaneous equation (2):
"OP" may have overlooked the fact, that since
there are four unknown's (a,b,c,d) and there are
six equations. Hence there are more equations than
unknown's. Hence the equations becomes redundant.