Solve the following equations (integer solutions )

Question

$$ax + p = 1341$$ $$by - q = 1341$$ $$x+y = p+q$$ $$6(x+y) =a+b$$ $$ar-p = 6710$$ $$x+r=b$$ I'm Looking for integer solutions for all the variables with the constraint a,b>0. I could reduce this equation further and eliminate a few variables but I am wondering if these equations can be solved efficiently without having to factor 8051. (1341+6710)Thanks

Answer 1

Substituting $x,p,q,r,b$ from the equations, only one equation is left, namely the third one, $$ (a^2 + 8051)(a - 6y + 1)=0. $$ This implies $y=(a+1)/6$, and all solutions are given by $$ (a,y,x,p,b,q,r)=\left(a,\frac{a+1}{6},\frac{8051-a}{6a}, \frac{a-5}{6}, \frac{8051}{a},\frac{5a + 8051}{6a},\frac{a + 40255}{6a}\right). $$ For integral solution we need that $a$ is a divisor of $8051$ such that $a\equiv -1 \bmod 6$. This gives the two positive solutions $a=83$ and $a=8051$, and hence the two integral solutions $$ (a,y,x,p,b,q,r)=(83,14,16,13,97,17,81), $$ and $$ (a,y,x,p,b,q,r)=(8051,1342,0,1341,1,1,1). $$