Proof of Theorem 3.2 Since and are polynomials in with no common factors, Lemma 2.2 gives us
On the other hand, by the definition of , we get from (2.2)
and from Lemma 2.1
Combining (4.1) and the last two inequalities, we obtain
which is , then from (3.2). □
Proof of Theorem 3.3 We restrict p and q duo to the reasoning by Grammaticos et al. , where and are polynomials in w with constant coefficients.
Applying Lemma 2.2 to (1.2), we get from Lemma 2.5
Then . Rewriting (1.2) gives
Since and have no common factors, the right side of (4.2) is irreducible. Applying Lemma 2.2 now to (4.2), we get from Lemma 2.5 and Lemma 2.7
so . Thus . If , the degree of that was denoted by k would be 5 since , a contradiction. Hence, , , and .
If , since the degree of , , we have and the coefficients of the highest degree of P and Q are identical. □
Proof of Theorem 3.4 We get from (3.3)
where , , and . Then , , , and Λ of equals 4. It follows from Theorem 3.2 that
thus . In a similar way to the proof of [, Lemma 4.2], we have
for . From Lemma 2.6, given , there are at most points where , but where has a pole of order greater that or less than due to poles or zeros of . The combined effect of all such points can be included in the error term, and so we only consider the rest of the zeros of Q in what follows.
For a point where , define
to be the longest possible list of points such that each is a zero of , , and each is a pole of w.
Suppose that w has more than poles that are not contained in any sequence . Let be the integrated counting function of such poles; by assumption we have for some in a set of infinite logarithmic measure. By (3.3), , and so we get
which implies that by Lemma 2.3. Therefore all except at most poles of w are in some sequence .
We will call the total number of zeros of in divided by the total number of poles of w (both counting multiplicities) the ratio of the sequence.
Consider a sequence that contains only one zero of . Then there are one or two poles in that sequence. With one pole we would have and or , where . If there are two poles, the situation is the same except that now we have . In any case, in such a sequence the ratio is at most . We suppose that there are more than , contains only one zero of . Then there are more than sequences such that . Hence, we have
which is a contradiction, thus all except at most sequences contain at least two zeros of . This means that there must be at least points such that and one of the following holds:
Since holds for both choices of , exactly one of the following is true:
Both (4.7) and (4.8) hold .
Both (4.5) and (4.6) hold , (4.7) and (4.8) hold .
Relation (4.6) holds , (4.5), (4.7), and (4.8) hold .
Relation (4.5) holds , (4.6)-(4.8) hold .
In what follows, we will derive some consequences separately for the conditions (i)-(iv). We rewrite (3.3) as
where and .
Case (i) holds. Now we have both (4.7) and (4.8) hold . If (4.7) holds , we get from (4.9)
which means holds at more than points , then . Similarly, if (4.8) holds , we get . In this case, we have
Case (ii) holds. In the same way as above, we have
Cases (iii) and (iv) hold. Assume that (iii) holds. As we know, contains at least two zeros of . From (4.6), there are exactly one zero of and one zero of , otherwise, (4.5), (4.7), and (4.8) will hold . Then all except at most poles of w must be contained in sequences of the form
where if , the corresponding endpoint of the sequence is a zero of order , and if , it is some nonzero finite value. By Lemma 2.6,
holds for both choices of the ± sign. Denote
Next, we will show that U is a small function with respect to w. From (4.3), we get . From the definition of U and the fact that all but at most poles of w are in sequences of the above form, it follows that if U has more than poles, then there are more than sequences where .
For the sequences with , we may assume that for all such sequences, otherwise the poles with will only have a small effect (at most ) on . The ratio for the sequences in consideration is
using (4.12). Take d such that
Then . For a fixed j, there exists an satisfying
Define . Since
Noting that , then ϵ is well defined. Thus we conclude that if , then in such sequences the ratio is at most some . We will get a contradiction as the above. If , we will get a contradiction similarly. Therefore, , and so (3.3) becomes the Riccati difference equation
The same reasoning works for case (iv) as we exchange the roles of and .
In the case that , the proof is similar. The condition (i) is the only possibility. □
Restriction of . We claim that for each ,
for some . Following the same reasoning as in the proof of [, Lemma 4.2], we have
provided is not a solution of (3.3), i.e., does not satisfy . In the following, we may assume that has a large number of zeros, or (4.14) holds as desired.
Consider the points where . From (3.3) we get
for all except at most points and for either or both choices of ±.
We assume the condition (i) or (ii) is true, otherwise, w satisfies a Riccati difference equation. We discuss the following three cases:
Case 1. More than points satisfy
Then . By
we have . If for , then . It follows that (note the conditions (i) and (ii)). Then from
we get , which means
Combining the above equation with (4.10) or (4.11), we have for or , which is a contradiction.
holds for or .
Case 2. More than points satisfy
By a similar reasoning to the above, we obtain
holds for or .
Case 3. More than points satisfy
Also, more than points satisfy
In this case, we have both (4.16) and (4.17).
Note the above three cases, all except at most points are contained in one of forms the sequences of
where , , , and φ is a pole or a finite value but not the zero of . In fact, if φ is the zero of , it will be a starting point of another sequence.
If (4.20) holds for more than points, there are more than sequences such that , where . This is a contradiction. Hence, (4.20) holds at most points. Then almost all the zeros of are in the sequences (4.18) and (4.19).
However, noting that and , we have , which contradicts with Lemma 2.3.