Conjecture:
Given $a,b\in\mathbb N_+$ with $\gcd(a,b)=1$ there are no $m,n\in\mathbb N\ni 0$ such that $(a-1)(b-1)-1=ma+nb$, but such $m,n\in\mathbb N$ exists for all $N\geq(a-1)(b-1)$.
I guess it is possible to construct $m,n\in\mathbb N$ such that $(a-1)(b-1)=ma+nb$ but I can't see how right now, and that it is possible to proceed by induction, but how to prove that there is no $m,n\in\mathbb N$ such that $(a-1)(b-1)-1=ma+nb$? Thankful for all help!
I was looking for a binary operation $f(a,b)$ as above and noticed that $f(a,b)=f(b,a)$, $f(1,b)=1$ and $f(2,b)=b$ similar to cancelation and identity for multiplication and tried $f(a,b)=(a-1)(b-1)$ which seems to be correct due to some testing.
Example $a=4, b=9$ where $3\cdot 8=24$:
{0,4,8,9,12,13,16,17,18,20,21,22,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,...}
I could also need some help finding $m,n\in\mathbb N$ such that $(a-1)(b-1)=ma+nb$ and with the induction part.
