This question has two parts.

Part $1$:

I've been learning how to solve Linear Diophantine Equations (LDE) in 3 variables and so far, I've encountered two methods:

Method $1$:

Problem: Find positive integral solutions of $7x+11y+26z=123.$

Solution: The greatest coefficient is $26$ and $7+11+26z\leq 123,$ hence $z\leq 4.$ Therefore the only possibilities are: $$z=1, 7x+11y=97,\text{ of which there is one solution }(6,5),$$ $$z=2, 7x+11y=71,\text{ of which there is one solution }(7,2),$$ $$z=3, 7x+11y=45,\text{ of which there is no solution},$$ $$z=4, 7x+11y=45,\text{ of which there is no solution}.$$ Thus there are two solutions in positive integers, namely $(6,5,1)$ and $(7,2,2)$.

Method $2$:

Problem: Determine integral solutions of $3x+4y+5z=6.$

Solution: We have $3x+4y\equiv 1 \pmod 5;$ hence $3x+4y=1+5s$ for some integer $s.$ Observe that $x=-1+3s, y=1-s$ satisfies this equation and therefore a general solution is $x=-1+3s+4t,y=1-s-3t$ for some $t.$ Substituting these values in original equation we get $z=1-s.$

Doubts regarding the above methods:

  1. Is Method $1$ a general way to solve LDEs in 3 variables or is it just a trick? Moreover, I noticed that it does not help one to solve the second LDE since $3+4+5z\leq 6\Rightarrow z\leq -0.2$ which basically means that we have no value of positive $z$ satisfying the equation. Does that mean that the second LDE does not have any positive integral solutions?
  2. How can we impose integral bounds on the parameters $s$ and $t$, if we want to obtain positive solutions? I tried to do the following: $$x\geq 0\Rightarrow -1+3s+4t\geq 0\Rightarrow 3s+4t\geq 1$$ $$y\geq 0\Rightarrow 1-s-3t\geq 0\Rightarrow s+3t\leq 1$$ $$z\geq 0\Rightarrow 1-s\geq 0\Rightarrow s\leq 1$$ Beyond this, I could not deduce anything meaningful.

Part $2$:

After having learned about the solvability of such equations, a natural question that comes to one's mind is: how many positive integral solutions does an LDE in three variables admit, for a given output. So suppose we have an LDE in three variables $$ax+by+cz=n,\text{ where }\gcd(a,b,c)|n$$ then how many triples $(x,y,z)$ exists such that $x,y,z\geq 0?$ I guess this is related to the last question in Part $1$ because if we can bound the parameters then we can simply get the number of solutions by counting the values satisfied by $s$ and $t$.


Method 1) is not a trick: it is the projection of the equation plane onto a 2D plane. To do that efficiently, you choose the variable (suppose $z$) with the highest coefficient, assign each possible value to it, and project it as line onto the plane of the remaining variables ($x,y$).
To determine the range of possible values for $z$, since you are requiring positive solutions (thus excluding the $0$) , you assign $1$ (the lowest) to $x$ and $y$ and then you get an inequality as in the example you gave.

Applying this method to the second example ($3x+4y+5z=6$) you correctly deduce that the possible range for $z$ is empty, since there are no positive solutions to this equation.

To my knowledge, unfortunately there is not a general "formula" to solve 3D LDE (not a simple one at least), and also algorithms to calculate them become computationally heavy for large coefficients.


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