Consider the Diophantine equation $ \ px+qy=n \ $ , where $ \ p , q \in \mathbb{Z}^+ $ and
$ \ n \in \mathbb{Z}^- $.
Let $ \ \gcd(p,q) \ $ be a factor of $ \ n \ $, i.e. $ \ \gcd(p,q) \ =\ nk $ with $k\in\mathbb{Z}$
Show that if $ \ n >pq-(p+q) \ $ then $ \ px+qy=n \ $ has at least one solution in positive integers .
Answer:
Let $ \ \gcd(p,q)=d \ $
Then $ \ d \ |n \ $ and which implies $ \ n=kd \ $ for some $ \ k \in \mathbb{Z} \ $.
Further since $ \ \gcd(p,q)=d \ $ , there exists $ \ x' , y' \in \mathbb{Z} \ $ such that
$$ \ px'+qy'=d. \\ \text{Multiplying by $k$ gives} , \\ k(px'+qy')=kd \\ \Rightarrow p(kx')+q(ky')=n \ $$
Let $ \ x_0=kx', \ y_0=ky' \ $ , then
$ px_0+qy_0=n \ $ giving solution $ \ (x_0,y_0) \ $
But how to use the given condition in order to get positive integer solutions i.e. $ \ x_0 , \ y_0 \ $ will be positive integer ?
Help me out
You should look up the Euclidean algorithm: https://en.wikipedia.org/wiki/Euclidean_algorithm#Worked_example You can use this to determine the values in the formula of Bézout, that is $\gcd(p,q)=px+qy=:d$ with $x,y\in\mathbb{Z}$.