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# Theorems of Special Interest

Lemma 1   : , for all .

Proof: Let . Then

Thus the statement is true for the base case. Now assume that the statement is true for some and let . We must show that

By assumption, , so we have

as required. Thus we conclude by induction that the statement is true for all .

Theorem 1   : Let . Then if and only if

Proof: Assume that . By Lemma 1, . But , so as required. Conversely, assume that . Again by Lemma 1, so . Thus and we are done.

For the remainder of this report, let and let denote the conjugate of .

Theorem 2   : The continued fraction expansion of is purely periodic if and only if .

Proof: Suppose that the continued fraction expansion of is purely periodic. Then this expansion is of the form . Let . Then by Theorem 1,

Rearranging terms we obtain the quadratic equation . Solving for x we have

 (6)

But since det , so . Substituting in (6) we have

Since , . Thus

By the construction of , , , , , , and . Thus . By properties of convergents, . Thus rewriting as we see that

It follows that .

(This proof of the converse, and an alternate, more precise proof of the above, can be found in [2])
Conversely, suppose that and . Let and recursively define

Since , for all . Therefore, if , then and . Because , it follows by induction that for all . Thus we have that

Since is a quadratic irrational, it must be that for some with . Then and

It is also true that

Thus implies that . Doing this times yields

as required.

Let be the greatest integer in and let . Then obviously is larger than and is strictly between and . Thus applying the previous theorem, the continued fraction expansion of must be purely periodic. In fact we see that

 (7)

Applying Theorem 1, we see that if and only if . Thus we have the following corollary to Theorem 1 (in matrix notation).

Corollary 1 (to Theorem 1)   : Let be a square-free integer whose continued fraction representation has a center, be the greatest integer in , as defined in (4), and . Then if and only if

The following proposition will also be useful. A proof can be found in Justin Miller's report: Families of Continued Fractions.

Proposition 1   : Let . Then if and only if

where , and is a fractional linear transformation.

Using this proposition it follows that if , then

Solving for yields

 (8)

Next: Proof of Main Theorem Up: Continued Fraction Factorization Previous: Matrix Representations and Fractional

scanez 2000-12-04