Greatest common divisor of $3$ numbers Let $a,b, c$ belong to $\mathbb Z$ such that $(a,b,c) \neq (0,0,0)$. Define the
[highest common factor] greatest common divisor ${\rm gcd}(a, b, c)$ to be the largest positive integer that divides $a, b$, and $c$.  

Prove that there are integers $s, t, u$, such that
  $${\rm gcd}(a, b, c) = sa + tb + uc$$
  Find such integers $s, t, u$ when $a = 91, b = 903, c = 1792$.

 A: Below is one conceptual way to prove Bezout's Identity for the gcd = hcf. For computation of the Bezout identity it is convenient to employ the extended Euclidean algorithm.
The set $\rm\,S\,$ of integers of form $\rm\,a\,x + b\,y+c\,z,\,\ x,y,z\in \mathbb Z,\,$ is closed under subtraction so, by the Lemma below, every positive $\rm\,n\in S\,$ is divisible by $\rm\,d = $ least positive $\rm\in S.\,$ So $\rm\,a,b\in S$ $\Rightarrow$ $\rm d\mid a,b,\,$ i.e. $\rm\,d\,$ is a common divisor of $\rm\,a,b,\,$ and greatest: $\rm\ c\mid a,b\,$ $\Rightarrow$ $\rm\,c\mid d = a\,x\!+\! b\,y\!+\!c\,z\,$ $\Rightarrow$ $\rm\,c\le d.$   
Lemma $\ \ $ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\  n-m\, \in\, S.\,$  Then every element of $\rm\,S\,$ is a multiple of the least element $\rm\:\ell = \min\, S.$
Proof ${\bf\ 1}\,\  $ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra  $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$
Proof ${\bf\ 2}\,\rm\,\ \   S\,$ closed under subtraction $\rm\,\Rightarrow\,S\,$ closed under remainder (mod), when it is $\ne 0,$ since mod may be computed by repeated subtraction, i.e. $\rm\, a\ mod\ b\, =\, a - k b\, =\, a-b-b-\cdots -b.\,$ Thus $\rm\,n\in S\,$ $\Rightarrow$ $\rm\, (n\ mod\ \ell) = 0,\,$ else it is $\rm\,\in S\,$ and smaller than $\rm\,\ell,\,$ contra mimimality of $\rm\,\ell.$ 
Remark $\ $ In a nutshell, two applications of induction yield the following inferences
$\ \ \rm\begin{eqnarray} S\ closed\ under\ {\bf subtraction} &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction \\
&\:\Rightarrow\:&\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm) \end{eqnarray}$
Interpreted constructively, this yields the extended Euclidean algorithm for the gcd. Namely, $ $ starting from the two elements of $\rm\,S\,$ that we know: $\rm\ a \,=\, 1\cdot a + 0\cdot b,\ \ b \,=\, 0\cdot a + 1\cdot b,\ $ we search for the least element of $\rm\,S\,$ by repeatedly subtracting elements of $\,\rm S\,$ to produce smaller elements of $\rm\,S\,$ (while keeping track of each elements linear representation in terms of $\rm\,a\,$ and $\rm\,b).\:$ This is essentially the subtractive form of the Euclidean algorithm (vs.  mod/remainder form).
