Largest integer that can't be represented as a non-negative linear combination of $m, n = mn - m - n$? Why?

Question

This seemingly simple question has really stumped me:

How do I prove that the largest integer that can't be represented with a non-negative linear combination of the integers $m, n$ is $mn - m - n$, assuming $m,n$ are coprime?

I got as far as this, but now I can't figure out where to go:

$mx + ny = k$, where $k = mn - m - n + c$, for some $c > 0$

$\Rightarrow m(x + 1) + n(y + 1) = mn + c$

If I could only prove this must have a non-negative solution for $x$ and $y$, I'd be done... but I'm kind of stuck.

Any ideas?

Answer 1

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 $(m-1)n-m = mn-m-n$.

Answer 2

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.

Answer 3

We give two proofs below - first a purely arithmetical proof, then a more geometric version.

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First we consider positive solutions $\,x,y\ge 1,\,$ then we generalize that to $\,x,y\ge c\,\,$ by a shift. Below we use these well-known results: $\,\gcd(m,n)=1\,$ $\Rightarrow\,\color{#90f}{m^{-1}\ \rm exists}$ $\!\bmod {n}\,$ (e.g. by the Bezout gcd identity) and furthermore $\,m\mid nk\color{#4bf}{\overset{\!\rm \small E}\Rightarrow} m\mid k\:$ (by $\, \color{#4bf}{\rm \small E} =$ Euclid's Lemma).

$ mx+ny = \color{#0a0}{k > mn}\,$ is $\rm\color{#0a0}{solvable}$ for $\,x,y\ge 1,\,$ by mod $n\!:\, $ $\rm\color{#90f}{there\ is}$ an $\,x\equiv \color{#90f}{m^{-1}}k\,$ with $\, 1\le \color{darkorange}{x \le n},\,$ so $\, mx \equiv k,\,$ so $\,m x + n y = k,\,$ for $\, y\in\Bbb Z,\,$ and $\,y>0\,$ by $\,m\color{darkorange}x \le \color{#0a0}m\color{darkorange}{n}\color{#0a0}{< k}$

$ mx+ny\: \color{#c00}{{\bf =}\: mn}\,$ is $\rm\color{#c00}{unsolvable}$ by $\, m\:\!|\:\! ny\ \smash[t]{\color{#4bf}{\overset{\rm \small E}\Rightarrow}}\ m\:\!|\:\!y\ $ so $\ x+n(y/m) = n\,$ contra $\,x,y/m \ge 1$

Remark $\ $ A simple shift translates the above to handle $\,x,y \ge c,\,$ viz. $\,\ \ \ \ \ \ \begin{align} &\ \ \ \ \, m\,x^{\phantom{|^|}} \ \:\!+\ \ \ \ n\,y\ \ \ \ =\ k\qquad\qquad\quad\ \:\!{\rm for}\ \ x,y \ge c = \bar c\!+\!1\\[.2em] \iff\ &m(x\!-\!\bar c) + n(y\!-\!\bar c) =\, k\!-\!\bar c(m\!+\!n)\,\ \ {\rm for}\ \ x\!-\!\bar c,\,y\!-\!\bar c\:\!\ge\:\! 1,\ \text{so by above}\\[.2em] &{\rm this\ \ is\ \underset{\textstyle\color{#c00}{unsolvable}}{\color{#0a0}{solvable}}\ for}\ \ \,k\!-\!\bar c(m\!+\!n)\underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}{\bf \color{#0a0}>} mn\ \ {\rm i.e.}\ \ k \underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}{\bf \color{#0a0}>} mn\!+\!\bar c(m\!+\!n)\\[.1em] \end{align}$

The bound $\, {\cal F}_c = mn + (c\!-\!1)(m\!+\!n)\,$ is known as the Frobenius number. The most common cases $ $ are: $\ {\cal F}_0 = mn-m-n;\:$ $\:{\cal F}_1 = mn.\,$ Note $\,{\cal F}_c\,$ is sometimes called modified if $\,c\neq 0.$

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Below is a more geometric proof that $\,{\cal F}_0 = mn-m-n$.

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.

Answer 4

Here's another version of the proof. Make a chart of the nonnegative integers in rows of size $m$, then mark the first $m$ multiples of $n$ (from 0 up to but not including $mn$). For example, if $m=4$ and $n=7$, we have the chart below with the first 4 multiples of 7 marked with a * :

\begin{array} & 0* & 1 & 2 & 3 \\ 4 & 5 & 6 & 7* \\ 8 & 9 & 10 & 11 \\ 12 & 13 & 14* & 15 \\ 16 & 17 & 18 & 19 \\ 20 & 21* & 22 & 23 \\ 24 & 25 & 26 & 27 \\ 28 & 29 & 30 & 31 \\ \ldots \end{array}

Now observe:

1. Every column has exactly one marked value. (This follows from (m,n)=1.)
2. The marked values, and all values below in the same column, are all representable as non-negative linear combinations of $m$ and $n$.
3. Conversely, any representable non-negative integer $mx+ny$ lies on or below the marked value in its column. (For $mx+ny$ must be $x$ rows below the value $ny$, which is a multiple of $n$ and therefore lies on or below a marked value.)

Therefore, the largest non-representable number lies one row above the largest marked number. This is $(m-1)n -m = mn - n -m $.

Answer 5

I think the easiest way to get the idea is as follows. Below are two basic facts that lead almost immediately to the answer. Assume that $m<n$ and $s=0\ldots(n-1)$.

1. If $nk+s$ is representable as a non-negative linear combination of $m$ and $n$, then $n(k+1)+s$, $n(k+2)+s$ etc. are representable as well.

2. If $nk+s$ is the least positive number of this form that is representable as a non-negative linear combination of $m$ and $n$, then $nk+s=mt$ for some positive $t$ (indeed, otherwise if $nk+s=mt+nu,u>0$, then we can subtract $n$, and $n(k-1)+s$ will still be representable).

Now, from these two facts: for each $s=0\ldots(n-1)$ we can find the least $t_s$ such that $mt_s\equiv s \mod n$, and if $t_{s'}$ is the largest among all $t_s$'s then $mt_{s'}-n$ is the largest number that cannot be represented as a non-negative linear combination. Since $m$ and $n$ are coprime, there is a one-to-one correspondence between $t_s\in\{1,\ldots,n-1\}$ and $s\in\{1,\ldots,n-1\}$, and $t_{s'}=n-1$ for $s'=n-m$. Hence, the largest non-representable number is $m(n-1)-n=mn-m-n$.