Euclid's GCD algorithm which is used to find GCD of two input numbers, say, $c$ and $d$, needs the inputs to be positive integers.

Exercise 12 provides an extension to this algorithm and allows $c$ & $d$ to accept values of the form $u+v\sqrt{2}$, where $u$ and $v$ are integers. In this case we can find a $r$ (of the form $u+v\sqrt{2}$) such that $c=dq+r$ , $q$ is a positive integer.
The algorithm can then continue as usual with $c$<-$d$ and $d$<-$r$ in the next step.

The algorithm will however not terminate if $c=1$ and $d=\sqrt{2}$ because there is no common divisor($q$) here.

However, the algorithm can be made to terminate in this case also if some extension is done to the divisor $q$, as explained here (by the author):

If we extend the concept of divisor so that $u+v\sqrt{2}$ is said to divide $a(u+v\sqrt{2})$ if and only if $a$ has the form $u'+v'\sqrt{2}$ for integers $u'$ and $v'$, there is a way to extend the algorithm so that it always will terminate. If we have $c=u+v\sqrt{2}$ and $d=u'+v'\sqrt{2}$, compute $c/d=c(u'-v'\sqrt{2})/(u'^2-2v'^2)=x+y\sqrt{2}$ where x and y are rational. Now let $q=u''+v''\sqrt{2}$ where $u''$ and $v''$ are the nearest integers to $x$ and $y$; and let $r=c-qd$. If $r=u'''+v'''\sqrt{2}$, it follows that $|u'''^2-2v'''^2|<|u'^2-2v'^2|$, hence the computation will terminate.

I did not understand the last line that

If $r=u'''+v'''\sqrt{2}$, it follows that $|u'''^2-2v'''^2|<|u'^2-2v'^2|$, hence the computation will terminate.

Please explain how $|u'''^2-2v'''^2|<|u'^2-2v'^2|$
and how this proves that computation will terminate.

  • 2
    $\begingroup$ To learn more about euclidean quadratic number fields I recommend having on hand at least one number theory textbook. This is discussed in many classical textbooks, e.g. those by Hardy & Wright, Harvey Cohn, and Harold Stark, to name just a few of many elementary expositions. $\endgroup$ – Bill Dubuque Dec 19 '14 at 16:58

The computation should terminate because of the same reason as the usual Euclidean algorithm, in which we search the greatest common divisor of natural numbers $c$ and $d$ with $c>d$. Namely, we have $c=qd+r$ with $r<d$. If $r>0$ then we have $d=q’r+r’$ with $r’<r$ and so forth. Following this way we consecutively construct a sequence $b>r>r’>r’’>\dots$ of non-negative integer residues, which have to be finite and stop at the zero. In the question to this sequence corresponds a sequence $|b|>|r|>|r’|>\dots$, where $|u+\sqrt{2}|=|u^2-2v^2|$. When the sequence stops at $|r_k|=|u_k+\sqrt{2}v_k|$ we have $|u_k^2-2v_k^2|=0$. Since $\sqrt{2}$ is irrational, the equality can hold only if $u_k=v_k=0$.

Algebraically, the proposed algorithm extension is a part of a proof that a ring consisting of numbers of the form $u+\sqrt{2}v$ is a Euclidean domain, endowed with a Euclidean function $|u^2-2v^2|$. In particular, that $|u'''^2-2v'''^2|<|u'^2-2v'^2|$ is proved in this answer.


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