Let a,b be coprime integers. Prove that every integer x>ab-a-b can be written as na+mb where n,m are non negative integers. Prove that ab-a-b cannot be expressed in this form.

We work under the given condtions, a, b positive coprime integers, gcd $(a,b)=1$. By Euclidean algorithm, we know that ANY integer x can be represented as $an+bm$ . The point is that we require that n and m be non-negative integers. To explain this, consider the example $a=10,b=3$. The statement says that every integer $x>30-10-3=17$ can be represented thus. Note that the integer 13 (less than 17) is also of this form , but 17 cannot be represented thus. We need to prove it in the general case.
Example: $a=10,b=3$.
The following are representable in the required form 3,6,9,10,12,13,15,16
The following are not representable in the required form 1,2,4,5,7,8,11,14,17.
But, as per theorm. all $x=18,19(x>17=ab-a-b)$ are representable in the required form.
Describing all integral solutions (no conditions) of the equation $an+bm=x$. The last line expresses all the solutions (n,m) in terms of a particular solution ${\displaystyle (\alpha ,\beta )}$ , we have crucially used that a and b are relatively prime.
Given: a,b positive integers, $(a,b)=1$
By Euclidean algorithm, ${\displaystyle \mathrm{\exists }\alpha ,\beta \in \mathbb{Z}}$, with ${\displaystyle a\alpha +b\beta =xZ}$
Consider any general solution
$an+bm=x$ Then ${\displaystyle an+bm=a\alpha +b\beta rAr\mathtt{a}(n-\alpha )=b(\beta -m)}$
But ${\displaystyle (a,b)=1\Rightarrow \mathrm{\exists }t\in \mathbb{Z}}$, with ${\displaystyle n-\alpha =tb,\beta -m=ta}$,
So, ${\displaystyle n=\alpha +tb,m=\beta =ta(t\in \mathbb{Z})}$
Thus, if ${\displaystyle \alpha ,\beta \in \mathbb{Z}}$ with ${\displaystyle a\alpha +b\beta =x}$, then all solutions $an+bm=x$, given by ${\displaystyle n=\alpha +tb,m=\beta =ta(t\in \mathbb{Z})}$
Note, that the m's can be brought to {0,1,2,...,a-1} In particular, m are all non-negative
We have shown that the largest x for which the representation $an+bm=x$ with n, m non-negative is not possible is $x=ab-a-b$. In other words, if $x>ab-a-b$, then x can be represented as $an+bm$ , for some non-negative integers a and b.
Now, note that ${\displaystyle \alpha <0\Rightarrow \alpha +tb<0\mathrm{\forall }t\le -\text{and}\beta -ta<0\mathrm{\forall }t\ge 0.}$
Thus, the set of all x not admitting non-negative solutions for $an+bm=x$ is ${\displaystyle \{a\alpha +b\beta :\alpha <0,\beta \in \{0,1,2,\dots ,a-1\}}$
The largest such x corresponds to ${\displaystyle \alpha =-1,\beta =a-1}$, with ${\displaystyle x=(-1)a+(a-1)b=ab-a-b.}$
We have already proved that x=ab-a-b is not representable in the required form. Here is another proof.
Now, the equation $an+bm=ab-a-b$, has a particular solution $n=-1,m=a-1$.
So, the general solution is $n=-1+tb,m=(a-1)-at$
${\displaystyle n\ge 0\Rightarrow tb\ge 1\Rightarrow t\ge 1}$
But ${\displaystyle t\ge 1\Rightarrow (a-1)-at<0}$
So $x=ab-a-b$ cannot be expressed as $x=an+bm$,n,m,non-negative integers.