Simple proof that there exist that you can obtain difference of 1 between multiples of coprimes $a$ and $b$ Given two coprimes $a$ and $b$ (assume wlog that $a < b$), there are non-negative integers $n_a$ and $n_b$ such that $n_b \cdot b = n_a \cdot a + 1$. Easy to prove using Bézout's identity, but is there a simpler, more "intuitive" way of proving this?
Another way of putting it is that there exists $n \in \mathbb{N}$ such that the remainder of the Euclidean division of $nb$ by $a$ is $1$.

$\gcd(a,b)=1$ implies $\bar b\in\mathbb Z/a\mathbb Z$ is invertible, so there exists $n_b\in\mathbb Z$ such that $n_bb\equiv1\pmod a$. Hence there exists $n_a\in\mathbb Z$ such that $n_aa=n_bb-1$, i.e. $n_bb=n_aa+1$.
