Sum of multiples of $2$ coprime numbers cover positive integers 
I have a conjecture that any two positive coprime integers $a$ and $b$ can be added to form any integer greater than or equal to $(a-1)(b-1)$.

This is true for $3$ and $5$, which can form any integer greater than or equal to $8$, where $8=3+5, 9=3+3+3, 10=5+5; k=3n+8$ or $3n+9$ or $3n+10$.
It is true for $4$ and $7$, which can form any integer greater than or equal to $18$, where $18=4+7+7, 19=4+4+4+7, 20=4+4+4+4+4, 21=7+7+7$, etc. How can it be proven that this will work for any two coprime numbers?
 A: You're interested in the set of numbers that can be written as $ax+by$ with $x,y \in \mathbb{N}$. So let $n \ge (a-1)(b-1)$ be an integer, we're interested in solving the equation $ax+by = n$ in $\mathbb{N}^2$.
Since $a$ and $b$ are coprime, you may find $u,v \in \mathbb{Z}$ such that
$$au + bv = 1$$
Multiplying the above equation by $n$, we get
$$aun + bvn = n$$
And for all $k \in \mathbb{Z}$ we have
$$a (un+kb) + b (vn-ka) = n$$
So the couples $(un+kb,vn-ka)$ are solutions in $\mathbb{Z}^2$ of our equation (and you may check that those are the only solutions in $\mathbb{Z}^2$, but we won't need that result). Let us prove that we can chose a value of $k$ such that $(un+kb,vn-ka)$ lies in $\mathbb{N}^2$ : Assume $a\ge 2$ (at least one of them is $\ge 2$, we just assume it's $a$), chose the smallest possible $k$ such that $un+kb \ge 0$, so its value is between $0$ and $b-1$ and we get
$$b(vn-ka) = n - (un+kb) \ge (a-1)(b-1)-(b-1) = (a-2)(b-1) \ge 0$$
which concludes. For more information you may want to look at  the Coin problem.
