Let a,b be coprime positive integers, find an integer N (depending on a and b), such that for any integer n > N it is possible to find integers s, t ≥ 0 satisfying sa+tb = n, but no such s, t exist satisfying sa+tb = N
I conjecture that N is ab-a-b, but I cannot prove it despite considerable efforts. Please help! Thanks.
Yes, $N=ab-a-b$.
Each of the numbers $0, a, 2a, \ldots, (b-1)a$ can have some nonnegative multiple of $b$ added to it so the sum is somewhere in the interval $[ab-a-b+1, ab-a]$. Because $a$ and $b$ are coprime, all these sums will be different. (Namely, if $sa+tb=s'a+t'b$ for $s\ne s'$, then $(s-s')a\equiv 0\pmod b$ and since $a$ is invertible modulo $b$ we have $s\equiv s'\pmod b$, but both $s$ and $s'$ are in $[0,b-1]$, so $s=s'$).
By the pigeonhole principe, then, every number between $ab-a-b+1$ and $ab-a$ will be hit, and every larger number can be made by adding an appropriate multiple of $b$.
It remains to show that $ab-a-b$ is not hit. However, if $sa+tb=ab-a-b$, then $s<b-1$ and $sa+(t+1)b=(b-1)a$ which contradicts the observation from above that each number in $[ab-a-b+1, ab-a]$ arises in only one way. So $N$ cannot be hit.