Consider the following linear Diophantine Equation::

```
ax + by + cz = d ------------ (1)
```

for all, a,b,c and d positive integers, and relatively prime, and assume a>b>c without loss of generality.

Can we find a lower bound on d which ensures at least one non-negative solution to this equation?

I know we can solve this problem easily for

```
ax+by = c. -------- (2)
```

The answer is

c>=ab, ------------(3)

which derives from the fact that the distance between two consecutive solutions of this equation is

```
$D = (\sqrt(a^2 + b^2))$ ----------(4)
```

and c>=ab ensures that the length of line in x-y plane is large enough to include at least one solution).

Since equation (1) is a plane in the xyz coordinate system, and the distance between consecutive solution can be shown to be DD = sqrt(b^2+c^2) (though this may not be smallest distance between solutions). I was thinking that if we can show that an inscribed circle with diameter DD can be enclosed within the triangle formed by x,y and z intercepts of Eq.(1), i.e. (c/a,0,0), (0,c/b,0) and (0,0,c/a), then we have at least one non-negative solution. But the in-circle radius has an inconvenient relationship with the original variables (a,b,c,d), and may not be a monotonic function of d.

Is there a smarter way to do this? and if such a bound exists, can it be extended to higher dimensions?

Thanks.