Let $a,b,c,d\in \mathbb{Z}$ and suppose we have the equation $ac+bd=1$.  One way of thinking about this equation is it expresses the fact $\gcd(b,d)=1$.  It is well-known that all other similar equations expressing that fact are of the form $(a+td)c+(b-tc)d=1$ for some $t\in\mathbb{Z}$.

Letting $a'=a+td,b'=b-tc$, one may ask if there is a "best" choice for $t$; or equivalently, if there is a best choice for $a',b'$.  In some sense, the usual Euclidean algorithm (applied to $b,d$) gives a minimal linear combination--we'll call the resulting pair $(a',b')$ the *Euclidean pair*.  But there are sometimes other situations where a different choice of $t$ is optimal.  In any case, we can pass from $ac+bd=1$ to $a'c+b'd=1$.

Now, we can view this latter expression as a linear combination showing that $\gcd(a',b')=1$.  So we can repeat the process to get a new $b',d'$ so that $a'b'+c'd'=1$.  Continuing in this fashion by replacing either the $(a,c)$ pair or the $(b,d)$ pair with a new pair $(a',c')$ or $(b',d')$ respectively, then after a finite number of steps we can always reach the equation $1\cdot 1+ 0\cdot 0=1$.  (One way to do this is just use the Euclidean pair for each replacement, except possibly at the end when one may need to pass from $0\cdot 0+1\cdot 1=1\mapsto 1\cdot 1+0\cdot 0=1$.)

My question is: Starting with any quadruple of integers $a,b,c,d\in\mathbb{Z}$ with $ac+bd=1$, is there an absolute bound $n\gg 0$ so that by performing some sequence of back-and-forth switching as described above we get to $1\cdot 1+0\cdot 0=1$ within at most $n$ steps?

The greedy algorithm of simply choosing the Euclidean pair will not give an absolute bound $n$, since we can easily construct a sequence which requires an arbitrarily large number of steps this way.  But using the Euclidean pair is not always the best choice.

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By the way, this number theory question originally arose in my study of perspective decompositions of the abelian group $\mathbb{Z}\oplus \mathbb{Z}$.  The connection comes from associating nontrivial idempotents $E\in \mathbb{M}_2(\mathbb{Z})$ with 4-tuples $(a,b,c,d)$ satisfying $ac+bd=1$, by writing $E=\begin{pmatrix}ac & bc\\ ad & bd\end{pmatrix}$.  [Here, $a$ is the gcd of the first column, etc...]

  Further, I have a very *ad hoc* way of showing that the question above has a negative answer if we replace $\mathbb{Z}$ with the ring $\mathbb{F}_2[x]$.  In that case, one can show that using Euclidean pairs *is* optimal.