Prove that $\exists c$ that the equation $ax+by=c$ has exactly n different positive solutions Let $a$ and $b$ same sign, $a,b\in \mathbb{Z}$ and $n\in\mathbb{N}$. Prove that $\exists c$ that the equation $ax+by=c$ has exactly n different positive solutions, it means $(x,y)$, where $x>0$ and $y>0$.
I have to use theorem: 
Let $a,b,c\in\mathbb{Z}$. If at least one of the numbers $a$ and $b$ is not $0$ and $x_0, y_0$ is the equation $ax+by=c$ some solution, then all solutions $x,y$ of this equation is obtained by means of formulas
$x=x_0+\frac{b}{gcd(a,b)}t$,
$y=y_0-\frac{a}{gcd(a,b)}t$,
giving all integer values to the variable $t$.
First find a solution $x_0,y_0$ and then suitable $c$ value.
How to prove it? 
Thank You.
 A: $a x+b y =c$
1- We find gcd of a and b: gcd(a, b)
2- the condition for existence of solution is that : $gcd(a, b) | c$
3- we find the solutions of homogeneous equation  $(ax+by=0)$; in case a and b are co-primes then $gcd(a, b)=1$ and we have: 
$a x+b y = 0$:
$x= b$ and $y=-a$
So general solutions of homogeneous equation is:
$x = b . t$
$y = -a . t$
Where t can be any arbitrary value in Z. If we sum these solutions with a couple of certain solutions $x_0$,and $y_0$ then we find  general solutions of equation $a x+b y =c$ ; we have:
$x = b.t + x_0$ 
$y = -a.t + y_0$
Hence this equation has infinitely many solutions in Z in general and in N in particular.The values of $x_0$ and $y_0$ depends on the value of c, therefore for n different c there can be n different sets of positive solutions.
Note that if a and b are co primes then their common divisor is $1$.
Geometric interpretation: Equation $ax +by=c$ in fact represent a line which is the locus of infinitely many points, among them many have coordinates in Z in general and in N in particular. With certain values of a and b , which define the slope of the lines c can have various values; each value gives a line parallel to others so that if we have n different c we will have n different equation or line with the same slope, hence we have exactly n solutions(or n lines) for the equation in each there exist many points having coordinates in N.So with n different c we have n different set of point having positive coordinates. c can be  any arbitrary integer. in other words multiplying both sides of equation by a common divisor like $d$ we get a new equation( a new line).In this way all equations satisfy the condition that $gcd(ad, bd)=d|cd=c_1$ where $c_1$ is the new c. 
A: Let $d=\gcd(a,b)$ and $a'=a/d$, $b'=b/d$. Then


*

*$x = x_0 + b't >0$ iff $t > -x_0/b'$

*$y = y_0 - a't <0$ iff $t < y_0/a'$.
So, $t \in (-x_0/b', y_0/a')$. The length of this interval is $$
\frac{y_0}{a'}+\frac{x_0}{b'}=\frac{a'x_0+b'y_0}{a'b'}=\frac{c}{a'b'd}=\frac{c}{m}
$$
where $m=lcm(a,b)$. Take $c=nm$ and the interval will have length $n$ and so $n$ integers inside it in general. If one or both extremes are integers, you'll have to compensate.
