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Theorem **Up:** Continued
Fraction Factorization **Previous:** Matrix Representations and Fractional

**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

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

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

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Theorem **Up:** Continued
Fraction Factorization **Previous:** Matrix Representations and Fractional