How can I solve (nontrivially) this equation in nonnegative integers: $$ x - 2y + 3z - 4t = 0.$$

By inspection I found the set of solutions is: {(2,1,0,0),(4,0,0,1), (0,3,2,0), (1,0,1,1), (0,0,4,3), (0,1,2,1), (1,2,1,0),(6,1,0,1)}

1- Is my solution correct?

2- Also, I got a hint that this can be solved as a linear Diophantine equation but I only know how to do this for 2 variable and sometimes for 3 (where I express the third variable in terms of the other two)but not for 4 variables.

Any help will be appreciated.

  • $\begingroup$ How did you miss the easiest one: $x,y,z,t=0$? $\endgroup$
    – vitamin d
    Feb 22 at 1:17
  • $\begingroup$ I am searching for nontrivial ones @vitamind $\endgroup$
    – user889696
    Feb 22 at 1:18
  • $\begingroup$ @hardmath I edited my post. $\endgroup$
    – user889696
    Feb 22 at 1:27
  • $\begingroup$ Are you saying applying Diophantine 2 times?@hardmath $\endgroup$
    – user889696
    Feb 22 at 1:28
  • $\begingroup$ I don't think I said "applying Diophantine 2 times" but you might be looking at the two "free parameters" $y,t$ identified in the Accepted Answer as giving us solution families that can be added to generate any possible solution. $\endgroup$
    – hardmath
    Feb 22 at 1:31

Clearly there are infinitely many solutions; scaling any solution yields another solution.

As for explicit solutions;for any $y,t\geq0$ and any $z\geq0$ such that $3z\leq2y+4t$ you have $$x:=2y-3z+4t\geq0.$$


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