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$$13x + 12 = 22y + 19 = 20z + 3$$


I am able to solve the first diophantine equation $13x + 12 = 22y + 19$ in the classic form of $ax + by = c$ like $$13x_1 - 22y_1 = 7.$$ In this case the condition of existing solution is met: $gcd(a, b) | c \rightarrow gcd(13, -22) | 7$. The solution is $x_1 = -35 + 22n, y_1 = -21 + 13n$.

I am also able to solve the equations:

  • $13x_2 + 12 = 20z_1 + 3$. The solution is $x_2 = 27 + 20n, z_1 = 18 + 13n$.
  • $22y_2 + 19 = 20z_2 + 3$. The solution is $y_2 = -8 + 10n, z_2 = -8 + 11n$.

Based on that, it is possible to "merge" solutions $x_1 = -35 + 22n$ and $x_2 = 27 + 20n$ into $x$ that fullfils the original equation?

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It is possible. If we get $z_1=z_2$ then it will apply $$13x_2+12=20z_1+3\ \ and \ \ 22y_2+19=20z_1+3$$ which is equivalent to your equation. Solution to $z_1=z_2$ is $$18+13n=-8+11k$$ $$n=-2+11m,\ k=13m$$ Where $n$ is parameter in solution of first and $k$ parameter in solution of second equation. Now substituting these into equations for $x_2,y_2,z_2$ we get:

  • $x_2=-13+220m$
  • $y_2=-8+130m$
  • $z_2=-8+143m$

which solves your equation.

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  • $\begingroup$ Is it possible aply this approach for more equations? For example: $13x + 12 = 22y + 19 = 20z + 3 = 31u + 12$, I am only interested in the variable $x$, respectively all results of $13x + 12$. $\endgroup$ – user8806343 Nov 5 '17 at 11:17
  • $\begingroup$ It is possible, you only have to solve $13x+12=31u+12$ to get $(x_3,u_3)$ then set $x_3$ and $x_2$ equal. This will yield solution to new equation $\endgroup$ – M C Nov 5 '17 at 12:48

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