Largest integer that can't be represented as a non-negative linear combination of $m, n = mn - m - n$? Why?
Here's an alternative (but perhaps more pedestrian) proof:
(a) $mn-m-n$ is not a non-negative linear combination: Assume, to the contrary, that $mn-m-n=am+bn$ with $a,b\in\mathbb N_0$. Then $$(a+1)m+bn=mn-n=(m-1)n$$ $$(a+1)m = (m-1-b)n$$ But $(a+1)m$ is clearly positive and since $(a+1)m < mn-n < mn$, it is a positive number less than $mn$ that is a multiple of both $m$ and $n$, contradicting the assumption that that $m$ and $n$ are coprime.
(b) Every integer $k>mn-m-n$ is a non-negative linear combination. The $m$ numbers $0$, $n$, $2n$, ..., $(m-1)n$ represent all the different residue classes modulo $m$, so one of them must be congruent to $k$ modulo $m$. So $k=am+bn$ where $0\le b<m$, and we need to check that $a$ is non-negative. However, if $a$ is negative, $am+bn$ can be at most $(m-1)n-m = mn-m-n$.
Let $m$ and $n$ be positive and relatively prime. We show that $mn$ is the largest integer that cannot be represented as a positive linear combination of $m$ and $n$, that is, as $mx+ny$ where $x$ and $y$ are positive integers. We then deduce the corresponding result for non-negative linear combinations. There are simpler proofs, but the one below fits naturally towards the beginning of a course in elementary number theory.
The proof consists of two parts: (i) $mn$ cannot be represented as a positive linear combination of $m$ and $n$, and (ii) every integer greater than $mn$ can be expressed as a positive linear combination of $m$ and $n$.
Non-Representability of $mn$: Suppose to the contrary that $mn=mx+ny$ where $x$ and $y$ are positive. Then $mx=n(m-y)$. Note that $m$ divides $n(m-y)$ and $m-y$ is positive. Since $m$ and $n$ are relatively prime, it follows that $m$ divides $m-y$. This is impossible, since $m>m-y>0$.
Representability of all integers $>mn$: Let $w$ be an integer greater than $mn$. We show that $w$ is representable.
Since $m$ and $n$ are relatively prime, some integer linear combination of $m$ and $n$ is equal to $1$. By multiplying through by $w$, we can find integers $x_0$, $y_0$ such that $$mx_0+ny_0=w.$$
Now let $t$ be any integer. It is easy to verify that $$m(x_0-tn)+ n(y_0+tm)=w.$$ We will show that we can choose $t$ so that $x_0-tn$ and $y_0+tm$ are both positive. Then setting $x=x_0-tn$ and $y=y_0+tm$ will give us the desired representation.
We want to choose $t$ such that $tn<x_0$ and $tm>-y_0$. So we want to find $t$ such that $$-\frac{y_0}{m} <t < \frac{x_0}{n}.$$
To show that we can find such an integer $t$, we look at the difference $$\frac{x_0}{n}-\left(-\frac{y_0}{m}\right).$$ But $$\frac{x_0}{n}-\left(-\frac{y_0}{m}\right)=\frac{mx_0+ny_0}{mn}=\frac{w}{mn}>1.$$ Since the interval $$-\frac{y_0}{m} <t < \frac{x_0}{n}$$ has length greater than $1$, it contains at least one integer $t$. This completes the proof.
In the same way, we can show that if $w>kmn$, then the equation $mx+ny=w$ has at least $k$ positive solutions.
Representability using non-negative $x$ and $y$: It is easy to see that $w$ is representable using positive integers if and only if $w-m-n$ is representable using non-negative integers. It follows that $mn-m-n$ is the largest number which is not representable using non-negative integers.
We give two proofs below - first a purely arithmetical proof, then a more geometric version.
First we handle the case of positive solutions $\,x,y\ge 1,\,$ then we perform a shift to handle $\,x,y\ge c.\,$ Below we use that $\,\gcd(m,n)=1\,\Rightarrow\, m^{-1}$ exists $\!\bmod {n}\,$ (e.g. by the Bezout gcd identity) and, furthermore, that $\,m\mid nk\,\Rightarrow\, m\mid k\,$ (e.g. by Euclid's Lemma).
$ mx+ny = \color{#0a0}{k > mn}\,$ is solvable for $\,x,y\ge 1,\,$ by mod $n\!:\, $ there is an $\,x\equiv m^{-1}k\,$ with $\, 1\le \color{#c00}{x \le n},\,$ so $\, mx \equiv k,\,$ so $\,m x + n y = k,\,$ for $\, y\in\Bbb Z,\,$ and $\,y>0\,$ by $\,m\color{#c00}x \le m\color{#c00}{n}\color{#0a0}{< k}$
$ mx+ny \color{#c00}{\bf =} mn\,$ is $\rm\color{#c00}{unsolvable}$ by $\, m\mid ny\,\Rightarrow\, m\mid y\ $ so $\ x+n(y/m) = n\,$ contra $\,x,y/m \ge 1$
Remark $ $ A simple shift translates the above to handle $\,x,y \ge c,\,$ namely with $\,\bar c = c\!-\!1,$ $\,\ \ \ \ \ \begin{align} &\ \ \ \ \, m\,x^{\phantom{|^|}} \ \:\!+\ \ \ \ n\,y\ \ \ \ =\ \ \ \ k\qquad\quad\ \ \ \ \ \ \:\!{\rm for}\ \ x,y \ge c = \bar c\!+\!1\\[.1em] \iff\ &m(x\!-\!\bar c) + n(y\!-\!\bar c) =\, k\!-\!\bar c(m\!+\!n)\,\ \ {\rm for}\ \ x\!-\!\bar c,\,y\!-\!\bar c\ge 1,\ \text{so by above}\\[.1em] &{\rm this\ \ is\ \underset{\textstyle\color{#c00}{unsolvable}}{ solvable}\ for}\ \ \,k\!-\!\bar c(m\!+\!n)\underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}> mn\ \ {\rm i.e.}\ \ k \underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}> mn\!+\!\bar c(m\!+\!n)\\[.1em] \end{align}$
The bound $\, {\cal F}_c = mn + (c\!-\!1)(m\!+\!n)\,$ is known as the Frobenius number. The most common cases are $\,{\cal F}_0 = mn-m-n;\,$ $\,{\cal F}_1 = mn.\,$ It is sometimes called modified when $\,c\neq 0.$
Key Idea $ $ In the plane $\,\mathbb R^2,\,$ a line $\rm\,a\,x+b\,y = c\,$ of negative slope has points in the first quadrant $\rm\,x,y\ge 0\ $ iff its $\rm\,y$-intercept $\rm\,(0,\,y_0)\,$ is in the first quadrant, i.e. $\,\rm y_0 \ge 0\,.$ We can use an analogous "normalized" point test to check if a discrete line $\rm\,m\,x + n\,y = k\,$ has points in the first quadrant.
By above (or linear diophantine theory) the general solution $\rm\,(x,y)\,$ of $\,\rm mx+ny = k\,$ is obtained by adding to a particular solution $\,(x_0,y_0)\,$ arbitrary integer multiples of $\,\rm (-n,m).\,$ Doing so we can normalize any solution to one in "least terms", i.e. having the least possible value of $\rm\,x\in\Bbb N$.
Hint $\ $ Normalize $\rm\,k = m\, x + n\, y\,$ so $\rm\,0 \le x < n\,$ by adding a multiple of $\rm\,(-n,m)\,$ to $\rm\,(x,y)$
Lemma $\rm\ \ k = m\ x + n\ y\,$ for some integers $\rm\,x,\,y \ge 0\,$ $\iff$ its normalization has $\rm\,y \ge 0.$
Proof $\ (\Rightarrow)\ $ If $\rm\ x,\, y \ge 0\,$ normalizing adds $\rm\,(-n,m)\,$ zero or more times, preserving $\rm\,y \ge 0\,.\,$
$(\Leftarrow)\ \,$ If the normalized rep has $\rm\ y < 0,\,$ then $\rm\,k\,$ has no representation with $\rm\, x,\,y \ge 0\,\, $ since to shift so that $\rm\,y > 0\,$
we must add $\rm\,(-n,m)\,$ at least once, which forces $\rm\,x < 0,\,$ by $\rm\,0\le x < n.\ $ QED
Since $\rm\, k = m\ x + n\ y\, $ is increasing in both $\rm\,x\,$ and $\rm\,y,\,$ the largest non-representable $\rm\,k\,$ has normalization $\rm\,(x,y) = (n\!-\!1,-1),\,$ i.e. the lattice point that is rightmost (max $\rm\,x$) and topmost (max $\rm\,y$) in the nonrepresentable lower half $\rm (y < 0)$ of the normalized strip, i.e. the vertical strip where $\rm\, 0\le x \le n-1.\,$ Thus $\rm\,(x,y) = (\color{#0a0}{n\!-\!1},\color{#c00}{-1})\,$ yields that $\rm\, k = mx+ny = $ $\rm m\,(\color{#0a0}{n\!-\!1})+n\,(\color{#c00}{-1}) = $ $\rm mn\! -\! m\! -\! n\ $ is the largest nonrepresentable integer. $\ $ QED
Notice that the proof has a vivid geometric picture: representations of $\rm\,k\,$ correspond to lattice points $\rm\,(x,y)\,$ on the line $\rm\, n\ y + m\ x = k\,$ with negative slope $\rm\, = -m/n\,.\,$ Normalization is achieved by shifting forward/backward along the line by integral multiples of the vector $\rm\,(-n,m)\,$ until one lands in the normal strip where $\rm\,0 \le x \le n-1.\,$ If the normalized rep has $\rm\,y\ge 0\,$ then we are done; otherwise, by the lemma, $\rm\,k\,$ has no rep with both $\rm\,x,y\ge 0\,.\,$ This result may be viewed as a discrete analog of the fact that, in the plane $\,\mathbb R^2,\,$ a line of negative slope has points in the first quadrant $\rm\,x,y\ge 0\ $ iff its $\rm\,y$-intercept $\rm\,(0,\,y_0)\,$ lies in the first quadrant, i.e. $\rm y_0 \ge 0\,.$
Remark $\ $ There is much literature on this classical problem. To locate such work you should ensure that you search on the many aliases, e.g. postage stamp problem, Sylvester/Frobenius problem, Diophantine problem of Frobenius, Frobenius conductor, money changing, coin changing, change making problems, h-basis and asymptotic bases in additive number theory, integer programming algorithms and Gomory cuts, knapsack problems and greedy algorithms, etc.