Explanation of this solution involving Linear Diophantine Equations

For the question:

For what values of $c$ does $8x+5y = c$ have exactly one strictly positive solution?

The solution is this

So I have 3 questions.

I understand everything up until the part where it says that

If $x_0, y_0$ is to be the only positive solution, then $n = 0$ is the only value of $n$ to give a positive solution.

1) I interpret the above statement as, when $n = 0$, $x = x_0$ and $y = y_0$, and since $x_0$ and $y_0$ are positive, $x_0$ and $y_0$ is the only positive solution. Is my interpretation correct?

2) I don't really understand the significance of $n = 1$ and $n = -1$ and how from that, they can say that

$$c\in \{8x+5y \space | \space 1\le x \le 5, 1\le y \le 8\}$$

3) How would you approach this question?

2) If $n=0$ gives the only positive solution then $n=1$ and $n=-1$ must give a non-positive solution. You have $x = x_0+5n$. If $n=1$, then $5n$ is positive and $x_0$ is positive, so $x$ is positive. That forces $y$ to be non-positive, otherwise you have another positive solution. So $y = y_0 -8(1)$ must be non-positive. That is $y_0-8\leq 0$, or $y\leq 8$. Since we want $y0$ to be a positive integer, we must have $y\geq 1$, hence $1\leq y \leq 8.$
Then if $n=-1$, one of $x$ and $y$ must be non-positive. But $-8(-1)$ is positive so $y_0 -8(-1)$ is positive, so $x = x_0+5(-1)$ must be non-positive. So $x_0 \leq 5$. Since we want $x+0$ to be positive, we must have $x_0 \geq 1$, so $1\leq x_0 \leq 5.$
3) I would start by graphing several example of $8x+5y=c$ against the integer lattice, until the above made clear sense.
• Why specifically choose $n=1$ and $n=-1$? Why not $n=2$ or $n=-2$? – Alex Twin Dec 6 '17 at 0:22
• There's one positive solution. Moving the smallest possible distance on the line has to change the sign of either $x$ or $y$. $n=2$ is not the smallest possible distance. – B. Goddard Dec 6 '17 at 0:35
If $x_0, y_0$ is a positive solution, then plugging $n=0$ into the general solution works. Given that $x_0, y_0$ is to be the unique positive solution, no other value of $n$ substituted into the general solution can produce a positive pair $x,y$. Setting $n=1$ gives a positive value for $x$, so in this case $y$ must be negative. That is, $y_0 - 8 n \le 0$ or since $n=1$, we have $y_0 \le 8$. Similar reasoning gives $x_0 \le 5$.