Let a,b be coprime inegers. Prove that every integer x>ab-a-b can be written as na-mb where n,m where are non-negative inegers. Prove that ab-a-b connot be expressed ib this form.

In the following, a and b are coprime positive integers, Note that , by Euclidean algorithm, ANY integer x is an integral linear combination na+mb. The point is we seek sufficient condtions on the integer x ( namely $x>ab-a-b)na+nb=x$ with both n and m non-negative,For instance , consider $a=10,b=3$, clearly relatively prime positive integers . Now $x=10+3=13$ is a non-negative integral linear combination of a and b , but 14,16, 17 are not. The theorem asserts that whenver $x>10\cdot 3-10-3=17$,x can be expressed as $10n+3m$ , with n and m non-negative integers.
General solution of (1) is derived , using the fact that a and b are relatively prime.
Let ${\displaystyle x\in \mathbb{Z}}$
As gcda,b)=1, ${\displaystyle \mathrm{\exists }\alpha ,\beta \in \mathbb{Z}}$, with ${\displaystyle a\alpha +b\beta =x}$ (1)
Let (n,m) be any solution of (1): $an+bm=x$
Then ${\displaystyle an+bm=a\alpha +b\beta \Rightarrow a(n-\alpha )=b(\beta -m)}$
As $gcd(a,b)=1$, this ${\displaystyle \Rightarrow \mathbb{R}t\in \mathbb{Z}}$. with ${\displaystyle n-\alpha =bt\text{and}\beta -m=at\iff n=\alpha +bt,m=\beta -at}$(2)
1) Observe that from (2), any two solutions for m differ by a multiple of a . So there exists a unique solution for m which lies in among $\{0,1,2,...a-1\}.$
2) We have characterized when we can obtain non-negative solutions.
General solution for $an+bm=x$ (1)
${\displaystyle n=\alpha +bt,m=\beta -at,t\in \mathbb{Z}}$ (2)
Fix m for which ${\displaystyle m=\beta -at\in \{0,1,2,..a-1\}}$
Consider the corresponding solution (n,m).
Claim: ${\displaystyle \alpha \ge 0\iff n\ge 0,m\ge 0}$
Proof: If ${\displaystyle \alpha <-bt}$, then ${\displaystyle n-\alpha +bt<0\mathrm{\forall }t<0\text{and}m=m=\beta -at<0\mathrm{\forall }t\ge 0.}$
So, ${\displaystyle \alpha <0\Rightarrow }$ at least one ofm,n less than 0.
We have thus demonstrated that whenver x is an integer $>ab-a-b$, there exists non-negative integer solutions to the equation $x=na+mb$. This proves the first part of the problem.
We have shown ${\displaystyle \{x\in \mathbb{Z}:an+bm=x\Rightarrow n\text{or}\in m<0\}=\{x:x-an+bm:n<0,m\in \{0,1,2,\dots ,a-1\}}$
The maximum such value corresponds to in $n=-1,m=a-1$, that is, when $x=-1(a)+(a-1)b=ab-a-b$,
Proof: The second part of the problem: with $(a,b)=1$ , ie a and b are relatively prime implies that ab-a-b is not a non-negative integer linear combination of a and b.
Now, $an+bm=ab-a-b$, admits $n=-1,m=a-1$ as a particular solution. As above, the genaral solution is $n=-1+tb,m=(a-1)-ta$
Claim:${\displaystyle n\ge 0\iff m<0.}$
Proof: n${\displaystyle \ge 0\iff tb\ge 1\iff t\ge 1\iff ta\ge a\iff m=(a-1)-ta<0}$
Conclusion" ab-a-b is not a non-negative linearinegral combination of a and b.
Answer: 1) ${\displaystyle (a,b)=1,a,b>0,x>ab-a-b\iff \mathrm{\exists }n,m\in \mathbb{Z}}$