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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 $P_{12k-7}$.


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  • $\begingroup$ Did you try Gaussian elimination? $\endgroup$ – Alexey Burdin May 26 at 19:59
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    $\begingroup$ $pa^2+qab+rb^2+s=0 \implies (2 a p + b q)^2 - (q^2 - 4 p r) b^2 = -4 p s$. If $(q^2 - 4 p r) >0$, then this Pell equation with (possible) infinite set solutions $(a,b)$. If $(q^2 - 4 p r) <0$, then set solutions is finite. $\endgroup$ – Dmitry Ezhov May 26 at 20:42
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    $\begingroup$ System simplify as $(a,b,c,d)=(a,41 a - 108 d,8 a - 21 d,d)$ $\endgroup$ – Dmitry Ezhov May 26 at 21:08
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    $\begingroup$ I've just got $(\star)\iff (\star\star)\iff b=41a-108d$ and $c=8a-21d$. $\endgroup$ – mathlove May 27 at 6:03
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    $\begingroup$ Add condition $3bd-ac=4$ implies Pell equation for (a,d): $(9 d - 2 a)^2 - 2 a^2 = -1$. Then (a,b,c,d)=(29,1,1,11),(1136689,33461,38081,431211),... $\endgroup$ – Dmitry Ezhov May 28 at 12:01
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$\frac{a-b}{4} = \frac{2a-9c}{7} = 27d-10a = 9c-2b = \frac{27d-10b}{41} = \frac{3d-5c}{4} \tag{$(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.

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  • $\begingroup$ p.s. I didn’t overlook the "more equations than unknowns" fact: I just wanted to present all six possible pairings (e.g. $a$ with $b$, $a$ with $c$, etc.). $\endgroup$ – Kieren MacMillan May 27 at 17:30
  • $\begingroup$ By the way, this solution isn’t correct: $27d-10a = 27(3w)-10(w)=71w$. $\endgroup$ – Kieren MacMillan May 27 at 17:44

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