Suppose $~m~$ and $~n~$ are coprime and both of them are greater than one. Is it right that equation $~mx + ny = (m-1)(n-1)~$ has solutions over non-negative integers?
For example $~ (x,y) = (6,0) ~$ satisfies equation $~3x + 10y = 18~$.
Thanks in advance.
Yes, this equation has a (unique) solution in non-negative integers.
Suppose $m > n$. Looking at it modulo $n$, if that equation holds, we have
$$mx \equiv (m-1)(n-1) \equiv -(m-1) \pmod{n}.$$
There is a unique integer $x_0 \in [0,n-1]$ with $mx_0 \equiv 1-m \pmod{n}$. Then $(m-1)(n-1) - mx_0$ is a multiple of $n$, and we need to see that it is not negative. Since $m(n-1) \equiv -m \not\equiv 1-m \pmod{n}$, we know $x_0 \neq n-1$, hence $0 \leqslant x_0 \leqslant n-2$ and
$$(m-1)(n-1) - mx_0 \geqslant (m-1)(n-1) - m(n-2) = m - (n-1) > 0.$$