Largest Number not Expressible as Sum of Multiples of Coprime Integers

Theorem
Let $a, b$ be coprime integers, each greater than $1$.

Then the largest number not expressible as a sum of multiples of $a$ and $b$ is the number:
 * $a b - a - b = \paren {a - 1} \paren {b - 1} - 1$

Proof
First we show that $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.

$a b - a - b = s a + t b$ for some $s, t \in \N$.

Note that $t b \le s a + t b < a b - b = \paren {a - 1} b$.

This gives $t < a - 1$.

We also have $\paren {a - t - 1} b = \paren {s + 1} a$.

Hence $a \divides \paren {a - t - 1} b$.

Since $a$ and $b$ are coprime, by Euclid's Lemma:
 * $a \divides a - t - 1$

As $a - t - 1 > 0$, by Absolute Value of Integer is not less than Divisors:
 * $a \le a - t - 1$

which is a contradiction.

Hence $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.

Next we need to show that all numbers greater than $a b - a - b$ can be so expressed.

assume that $a > b$ and we split the numbers into two cases:

Case $1$: $x = a b - a - b + k$ for $1 \le k \le b$
For $k = b$, we have $a b - a = a \paren {b - 1}$.

Notice that for $1 \le k < b$ and $0 \le s \le b - 2$:

and we see that, by Absolute Value of Integer is not less than Divisors:
 * $a b - a - b + k - \paren {b - 1} a = k - b$

cannot be a multiple of $b$.

We claim that one of $a b - a - b + k - s a$ is divisible by $b$, where $0 \le s \le b - 1$.

Suppose not. Then we consider each of the remainders when dividing by $b$.

There are $b - 1$ remainders excluding $0$.

However we have a set of $b$ integers.

By Pigeonhole Principle, two integers must share the same remainder.

Suppose we have $a b - a - b + k - s_1 a$ and $a b - a - b + k - s_2 a$ both having remainder $r$ with $s_1 \ne s_2$:
 * $\exists p, q \in \Z: a b - a - b + k - s_1 a = p b + r, a b - a - b + k - s_2 a = q b + r$

Then:

Hence $b \divides \size {s_1 - s_2} a$.

By Euclid's Lemma, $b \divides \size {s_1 - s_2}$.

But $0 < \size {s_1 - s_2} \le b - 1$.

This contradicts Absolute Value of Integer is not less than Divisors.

Therefore one of $a b - a - b + k - s a$ must be divisible by $b$, where $0 \le s \le b - 1$.

We have shown that $s \ne b - 1$.

Hence for some $s$ with $0 \le s \le b - 2$, $a b - a - b + k - s a$ is a positive multiple of $b$.

This gives:
 * $a b - a - b + k = s a + t b$ for some $s, t \in \N$

as required.

Case $2$: $x > a b - a$
By Division Theorem:
 * $\exists q, r \in \Z: 0 \le r < b: x - \paren {a b - a - b + 1} = q b + r$

We have:

hence $q > 0$.

Moreover:
 * $x - q b = a b - a - b + \paren {r + 1}$

which falls into Case $1$.

We have shown that:
 * $\exists s, t \in \N: a b - a - b + \paren {r + 1} = s a + t b$

Hence:
 * $x = s a + t b + q b = s a + \paren {t + q} b$

as required.