Show that if $\gcd(a, b) = 1$ then $\forall n, n > ab~\exists \ x,y \in \mathbb{N} : n = ax + by$. The question is the exercise 27 Chapter 1 of the book Analytic Number Theory by Apostle:

For $a,b \in \mathbb{N} $ show that
a. If $\gcd(a, b) = 1$ then for every $n > ab$ there exist positive integers $x$ and $y$ such that $n = ax + by$.
b. If $\gcd(a, b) = 1$ there are no positive integers $x$ and $y$ such that $ab = ax + by$.    

I don't know how to solve this problem. By induction it fails to be approachable: For $n=1$, suppose $ab+1=ax+by$. But $ab+1=a(b-a)+a^2+1$ and $a^2+1=by?$  
 A: HINT: You can first prove: if $1 \leq n \leq a+b$, then there exist $x,y$ with $|x| \leq b$ and $|y| \leq a$ such that $ax+by=n$. 
This can be proved by considering the pair with minimum value of $|x|+|y|$. For example, if $x>b$ and $y<-a$, then the pair $(x-b,y+a)$ gives a smaller value.
Now this can be used to prove (a). For example, if $aX+bY=1$, with $X<0$, and $|X|<b$, then $ab+1=a(X+b)+bY$.
A: If $(x_0,y_0)$ is any particular solution of the linear Diophantine equation $ax+by=n$ (such a solution is indeed exist since $\gcd(a,b)=1$), then 
$$
\left(x_0+\frac{b}{d}t\,,\,y_0-\frac{a}{d}t\right),
$$
where $d=\gcd(a,b)$ and $t$ is an arbitrary integer, are all its solutions. Now, we need to prove that there exist an integer $t$ such that 
$$
x_0+\frac{b}{d}t>0\quad\hbox{and}\quad y_0-\frac{a}{d}t>0
$$
that is 
$$
(\ast)\qquad-\frac{x_0d}{b}<t<\frac{y_0d}{a}.
$$
Since $ab>n$, then
$$
\frac{y_0d}{a}-\left(-\frac{x_0d}{b}\right)=d\cdot\frac{by_0+ax_0}{ab}=\frac{dn}{ab}>d\geqslant 1,
$$
so the interval $(\ast)$ defiantly contains a suitable integer, as required $\blacksquare$. 
