Finding 4 Integer Unknowns in terms of constants I posted this before, but explained it awfully so i'm trying again. This might be hard without explaining the problem i'm trying to solve, but i'm not sure that i could explain it clearly enough, or how much it would help.
Consider the equations:
$$k=wa+xe=yb+zf \\ f=b+d \\ a=c+e \\ k>0$$
Here $a,b,c,d,e,f$ are positive constant integers and: $$GCD(a,b,c,d,e,f)=1$$ $w,x,y,z$ are also positive integers.Given values for these 6 constants, I want to be able to find $w,x,y, z$ (or find a way to determine if they are odd or even) for the smallest possible value of $k$.
I would love any kind of help solving this, or even knowing if this problem is possible to solve with the current information, or if i need to look for a different way to go about this problem. I'm fairly sure the problem cannot be solved the way i'm trying to, but at the very least, I would like confirmation from someone more capable than me.
 A: (I see that there is now a new, recently added requirement that "$w,x,y,z$ are also positive integers." So this analysis is incomplete for the question as it is currently written: it was done before the last edit to the question. Consider this a partial analysis.)
Of your three equations, the second and third merely relate your constants so that $c$ and $d$, which are not even in the first equation, are determined from the other constants. So you really have just one equation relating your four variables. I'm sure you know that means that we can make a parameterization of three free real variables to give your four variables as real numbers. However, you have the added restriction that the variables are all integers. You also want $k$ positive and as small as possible. Those restrictions will make our parameterization have only two free variables with integer values.
Bézout's Identity with its corollaries says that the possible values of $wa+xe$, the left hand side of your equation, are precisely the multiples of the greatest common divisor of $a$ and $e$. Similarly, the possible values for $yb+zf$, the right hand side of your equation, are precisely the multiples of the greatest common divisor of $b$ and $f$. The smallest possible positive value that can be attained by both sides of your equation simultaneously is the least common multiple of those greatest common divisors. In other words, your desired $k$, the smallest possible value in your equation, is
$$k=\operatorname{lcm}(\operatorname{gcd}(a,e),\operatorname{gcd}(b,f))$$
It would be nice to simplify that, but your condition that $\gcd(a,b,c,d,e,f)=1$ does not tell us what the two gcd's in my formula are. So there is no simplification that I can see.
Now that $k$ is determined, we want to find all integral solutions to 
$$wa+xe=k, \qquad yb+zf=k$$
The process is too long to write here, but for given $a$ and $e$ we can use the Extended Euclidean algorithm to find integers $g$ and $h$ such that
$$ga+he=\operatorname{gcd}(a,e)$$
We could replace $g$ with $g+re$ and replace $h$ with $h-ra$ and the equality would still hold. Since $k$ is a multiple of $\operatorname{gcd}(a,e)$ we can multiply both sides of the equation by an integer to get $k$:
$$\frac{k(g+re)}{\operatorname{gcd}(a,e)}\cdot a
 +\frac{k(h-ra)}{\operatorname{gcd}(a,e)}\cdot e=k$$
Then, by a corollary to Bézout's Identity, if $r$ runs through the integers, that gives us all the solutions to $wa+xe=k$.
Similarly, we can use the Extended Euclidean algorithm to find integers $i$ and $j$ such that
$$ib+jf=\operatorname{gcd}(b,f)$$
And letting $s$ run through the integers, all solutions to $yb+zf=k$ are given by
$$\frac{k(i+sf)}{\operatorname{gcd}(b,f)}\cdot b
 +\frac{k(j-sb)}{\operatorname{gcd}(b,f)}\cdot f=k$$

In summary, given your restrictions, the smallest possible positive value of $k$ is

$$k=\operatorname{lcm}(\operatorname{gcd}(a,e),\operatorname{gcd}(b,f))$$

We use the Extended Euclidean algorithm to find constants $g,h,i,j$ (determined by the values of $a,e,b,f$) such that

$$ga+he=\operatorname{gcd}(a,e), \quad ib+jf=\operatorname{gcd}(b,f)$$

We get all solutions to your equation by choosing integral parameters $r$ and $s$ and setting

$$w=\frac{k(g+re)}{\operatorname{gcd}(a,e)}$$
  $$x=\frac{k(h-ra)}{\operatorname{gcd}(a,e)}$$
  $$y=\frac{k(i+sf)}{\operatorname{gcd}(b,f)}$$
  $$z=\frac{k(j-sb)}{\operatorname{gcd}(b,f)}$$


Here is a practical example using @Joffan's numbers $a=43,\ b=31,\ e=29,\ f=41$ ($c$ and $d$ are irrelevant):
$$k=1$$
$$g=-2,\ h=3,\ i=4,\ j=-3$$
$$w=29r-2,\ x=-43r+3,\ y=41s+4,\ z=-31s-3$$
A: You can ignore $a$ and $f$, and otherwise the answer - assuming positive $k$ - just seems to be the least common multiple of $b,c,d,e$. If $(b,c,d,e)$ are pairwise coprime, that is $k=bcde$.
OK, that was completely wrong, or (putting the best spin on it) only a distant upper bound. For a counter example to my first thought, take $(a,b,c,d,e,f) = (43,31,14,10,29,41)$ then $k=72$ and $w=x=y=z=1$.
There are in any case much better limits in less conveniently-arranged choices of numbers. Looking at $k=wa+xe$ and defining $g=\gcd(a,e)$ then when $g=1$, it is certain that any number above $ae-a-e = (a-1)(e-1)-1$ can be formed. If $g>1$, then any multiple of $g$ that is greater than $\frac {ae}g - a - e $ can be formed.
So overall if we also define $h=\gcd(b,f)$  then our largest $k$ is no larger than the first multiple of $gh$ greater than or equal to the maximum of $\left(\frac {ae}g - a - e, \frac{bf}h-b-f \right)$. Note that we have $\gcd(g,h)=1$ from the condition of $\gcd(a,b,c,d,e,f)=1$.
