# Solving a system of equalities in 4 variables (but no numeric constant)

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 $$P_{12k-7}$$.

• Did you try Gaussian elimination? – Alexey Burdin May 26 at 19:59
• $pa^2+qab+rb^2+s=0 \implies (2 a p + b q)^2 - (q^2 - 4 p r) b^2 = -4 p s$. 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. – Dmitry Ezhov May 26 at 20:42
• System simplify as $(a,b,c,d)=(a,41 a - 108 d,8 a - 21 d,d)$ – Dmitry Ezhov May 26 at 21:08
• I've just got $(\star)\iff (\star\star)\iff b=41a-108d$ and $c=8a-21d$. – mathlove May 27 at 6:03
• 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),... – Dmitry Ezhov May 28 at 12:01

$$\frac{a-b}{4} = \frac{2a-9c}{7} = 27d-10a = 9c-2b = \frac{27d-10b}{41} = \frac{3d-5c}{4} \tag{(1)}$$

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.

• 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.). – Kieren MacMillan May 27 at 17:30
• By the way, this solution isn’t correct: $27d-10a = 27(3w)-10(w)=71w$. – Kieren MacMillan May 27 at 17:44