# Further Results about a Special Fermat-Type Difference Equation

Authors: Hongwei Ma; Jianming Qi; Zhenjie Zhang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we prove the difference equationFz3+ΔcFz+c3=1 does not have meromorphic solution of finite order over the complex plane C. We also discuss an application to the unique range set problem.


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## Body

## 1. Introduction and Main Result

This kind of questions is from the analogues to the Fermat diophantine equations(1)fn+gn=1for a positive integern. When n>3, Gross [1] and Baker [2] proved that (1) does not nonconstant meromorphic solution in the complex plane C.Gross [1] and Baker [2] also showed for n>2 there are no entire solutions. For the case n=3, Gross [1] and Baker [2] also got the meromorphic solutions such as(2)fz=4−1/6P′z−11+3−1/241/8P=1/2+P′z/121/2Pand(3)gz=4−1/6Pz−11−3−1/241/8P=1/2−P′z/121/2P,whereP is a Weierstrass P-function.Now the equationw3+z3=l defines an algebraic function whose Riemann surface has genus 1, and there is accordingly a uniformization by Weierstrass elliptic functions. Weierstrass elliptic function Pz≔Pz,g2,g3 is a meromorphic function with double periods ω1,ω2 and defined as(4)Pz;w1,w2≔1z2+∑μ,ν∈Z,μ2+ν2≠01z+μw1+νw22−1μw1+νw22,which is even and satisfies, after appropriately choosingw1 and w2, to the form(5)P′z2=4P3z−1.In the same paper, Gross [1] conjectured all meromorphic solutions of f3+g3=1 are necessarily elliptic functions of entire functions. The conjecture was proved by Baker [2]. He proved.Theorem 1.
Any functionFG, which are meromophic in the complex plane and satisfy(6)F3+G3=1
have the form(7)Fz=fhz,Gz=ηfhz=ηf−hz=f−η2hz,
whereh is an entire function and η is a cube-root of unity.We assume that the reader is familiar with the standard notations and results such as the proximity functionmr,f, counting function Nr,f, characteristic function Tr,f, the first and second main theorems, lemma on the logarithmic derivatives etc. of Nevanlinna theory, (see [3, 4]). Given a meromorphic function f, we shall call a meromorphic function az a small function of fz if Tr,a=Sr,f, where Sr,f is used to denote any quantity that satisfies Sr,f=oTr,f as r→∞, possibly outside of a set r of finite logarithmic measure. Here, the order ρf is defined by(8)ρf=limsupr→∞logTr,flogr.Now, the estimate the order ofPz, Bank and Langley [5] indicates that(9)Tr,P=πAr21+o1,andρP=2.Nevanlinna value distribution theory of meromorphic functions has been extensively applied to resolve growth and solvability of meromorphic solutions of linear and nonlinear differential and difference equations (see [4, 6–10]).Below, we list some well-known facts concerning the order of composite meromorphic functions that can be found in Edrei and Fuchs [11], and Bergweiler [12].Theorem 2.
Letf be meromorphic and h be entire in C. When 0<ρf,ρh<∞, then ρf∘h<∞ and h is transcendental, then ρf=0.In recent years, Nevanlinna characteristic offz+c (c∈C\0), the value distribution theory for difference polynomials, Nevanlinna theory of the difference operator and the difference analogue of the lemma on the logarithmic derivative had been established, see e.g., [13–22]. Due to these theories, there has been a recent study on whether the derivative f′ or f can be replaced by the difference operator fz+c or Δcfz=fz+c−fz in the above question.We may want to study all meromorphic solutionsf of the following difference equation(10)fnz+fmz+c=1for a fixed nonzero constantc [23, 24]. Shimomura [25] proved that the (10) has an entire solution of infinite order for the n>m=1,. Later, Liu et al. [26] proved that Eq. (10) has no transcendental entire solutions of finite order when n≠m. If n=m=1, Liu et al. [27] proved that the solutions of (10) are periodic functions of period 2c, and proved that (10) has transcendental entire solutions of finite order for n=m=2.For a meromorphic functionsfz, we define its difference operators by(11)Δcfz=fz+c−fz,Δcnfz=Δcn−1Δcfz,n∈N,n≥2.Recently, Lü and Han [28] described a property of meromorphic solutions to equation in (6) with Gz≔Fz+c, for c∈C\0, as the following result.Theorem 3.
The difference equationFz3+Fz+c3=1 does not have meromorphic solution of finite order.It is natural to ask whether the shiftFz+c can be replaced by ΔcFz in above Theorem 3. In this paper, based on the ideas of [28], we mainly consider this problem and prove the following results.Theorem 4.
The difference equationFz3+ΔcFz+c3=1 does not have meromorphic solution of finite order.Proof.
It follows from (2) and (3) that		(12)Fz=1/2+P′hz/121/2Phz,Fz+c−Fz=η1/2−P′hz/121/2Phz
Then we have(13)Fz+c=η1/2−P′hz/121/2Phz+Fz=η+1/2+1−ηP′hz/121/2Phz.
Rewrite it as(14)η+1/2+1−ηP′hz/121/2Phz=1/2+P′hz+c/121/2Phz+c.
We know(15)P′hz3=2FzP′h−1.
AssumeρF<∞. Then, combining (13), (14) with (5), one has(16)4P3hz−1=12F2zP2hz−12FzPhz+3.
Finally, we get(17)P3hz=3F2zP2hz−3FzPhz+1.
So(18)3Tr,Phz≤2Tr,F+2Tr,Ph+O1,
andρPh<∞. Note that f is a of finite order and (9) and Theorem 2 combined force h to be a polynomial.
Notice whenPz0=0, then P′2z0=−1. By (5) Now, write all the zeros of Pz by znn=1∞ and |zn|→∞ as n→∞. Note that h is a polynomial, then that all the zeros of hz=zn are simple zeros of hz−zn for n>N, where N is a positive integer, and han=zn for n>N. Now, we divide two steps to prove our result.
Step 1. If there exists a subsequence of an (with an→a0, n→∞, which we still denote it by an) such that Pha0+c=0. Then we have P′ha0+c=±i and P′ha0=±i, so we also have a0 is a simple zero of Phz.
Differential (14), we have(19)1−ηP″hz3h′zPhz+c+η+1+1−ηP′hz3P′hz+ch′z+c=P″hz+c3h′z+cPhz+1+P′hz+c3P′hzh′z.
Substitutean (for enough large n) into the above equation and by Phan+c=0, Phan=0, we have(20)h′an+cP′han+cη+12+1−η2P′han121/2=h′anP′han12+P′han+c121/2.
Noting thatP′han=±i and P′han+c=±i. Without of loss generality, together with (14), we also assume there exists a sub-sequence of an (here we still denote it by an) such that four cases:
Case 1. If P′han=i and P′han+c=i, we get(21)η+1+1−η33ih′an+c=h′an1+3i3.
Case 2. If P′han=−i and P′han+c=i, we get(22)η+1−1−η33ih′an+c=−h′an1+3i3.
Case 3. If P′han=i and P′han+c=−i, we get(23)η+1+1−η33i−1h′an+c=h′an1−3i3.
Case 4. If P′han=−i and P′han+c=−i, we get(24)η+1−1−η33ih′an+c=h′an1−3i3.
Noting thathz,hz+c are polynomials and infinite many an (with |an|→∞ when n→∞), we would have to get(25)η+1+1−η33ih′z+c=h′z1+3i3,η+1−1−η33ih′z+c=−h′z1+3i3−1,η+1+1−η33i−1h′z=h′z1−3i3,η+1−1−η33ih′z+c=h′z1−3i3.
Eitherη=1 or η=−1/2±3/2i, and by comparing the highest coefficient of the both polynomials of the above equations, These are simply impossible.
So there exists a positive integerN satisfying Phan+c≠0 for n>N.
Step 2. In view of Phak=0 and P′2hak=−1, we have that Phan+c=∞ for n>N.
Thus, we get the proposition that the zeros ofPhz are the poles of Phz+c except for finitely many points. This is to say(26)Nr,1Phz=N¯r,1Phz+Mlogr≤N¯r,Phz+c+Mlogr,
whereM is a positive number. Noting that F is transcendental function, so Mlogr=Sr,F and (27)Nr,1Phz≤N¯r,Phz+c+Sr,F.
ByFz=1/2+P′hz/121/2/Phz and ρPh<∞, we have(28)Tr,F≤Tr,Phz+Tr,P′h+O1≤OTr,Ph,
which implies thatρF=ρPh and Sr,F=Sr,Phz. By the equation −G3=F3−1=F−1F−ηF−η2, we deduce that all zeros of F−1, F−η and F−η2 are with multiplicities at least 3.
Then, we have(29)2Tr,F≤N¯r,1F−1+N¯r,1F−η+N¯r,1F−η2+N¯r,F+Sr,F≤13Nr,1F−1+13Nr,1F−η+13Nr,1F−η2+Nr,F+Sr,F≤Tr,F+Nr,F+Sr,F,
which implies thatTr,F=Nr,F+Sr,F. It leads to mr,F=Sr,F=Sr,Phz.
Note the form ofF, we have 1/2/Phz=Fz−P′hz/121/2Phz. Furthermore, we have(30)mr,1Phz=mr,1/2Phz+O1≤mr,F+mr,h′zP′hzPhz+mr,1h′z+O1=Sr,Phz.
Then, all the above discussions yield that(31)Tr,Phz=Tr,1Phz+O1=mr,1Phz+Nr,1Phz+O1≤N¯r,Phz+c+Sr,Phz≤12Nr,Phz+c+Sr,Phz≤12Tr,Phz+c+Sr,Phz≤12Tr,Phz+Sr,Phz+OrρPh−1+ϵ,
a contradiction.
Thus, we finish the proof of Theorem4.Remark 1.
It is easy to get the solution of the equationFz+ΔcFz=1 is only for constant. For n=2, the solution of the equation F2z+Δc2Fz=1, Liu (see Proposition 5.3 in [19]) proved there is no nonconstant finite order entire solution.


## 2. An Application

For a meromorphic functionf and a set S∈C, we define(32)EfS=⋃a∈Szfz−a=0,countingmultiplicities.We say thatf and g share a set S CM, provided that EfS=EgS.In 1976, Gross [29] posed the following question:Question 1 (see [29, Question 6]).
Can one find two (or possibly even one) finite setsSjj=1,2 such that any two entire functions f and g satisfying EfSj=EgSjj=1,2 must be identical?
If the answer to Question1 is affirmative, it would be interesting to know how large both sets would have to be. Many authors have been considering about it, and got a lot of related results. Some of them are due to Yi [30], Mues and Reinders [31], and Frank and Reinders [32].
Recently, value distribution in difference analogues of meromorphic functions has become a subject of great interest [33, 34]. Zhang [35] obtained the following results.Theorem B.
LetS1=∞ and S2=1,ω,⋯,ωn−1, where w=e2π/ni and n is a positive integer, let c∈C. Suppose that f is a nonconstant meromorphic function of finite order such that EfzSi=Efz+cSii=1,2. If n≥4, then fz=tfz+c, where tn=1.Recently Lu¨ and Han [28] proved thatTheorem C.
Under the conditions of TheoremB and if n≥3, then the conclusion of Theorem B still holds.In this paper, we mainly consider thefz and Δcfz share the set which has three elements and get the following result.Theorem 5.
LetS1=1,η,η2, for η=−1/2±3/2i, and S2=∞. Take c≠0. Let f be a meromorphic function of finite order satisfying EfzSl=EΔcfzSl for l=1,2. Then, either fz=23fz+c or fz=23ηfz+c.Proof.
By Theorem C, we setF=f3. Obviously, Fz and ΔcFz share 1,∞ CM. Then we can assume that(33)ΔcFz−1Fz−1=eQz,
whereQ is an entire function. Note that F is of finite order, we have Q is a polynomial. Obviously, Tr,eQz=mr,eQz=Sr,F.
Rewrite (33) as(34)Fz+c=eQz+1Fz−eQz−1eQz+1,
(35)
F
z
=
1
e
Q
z
+
1
F
z
+
c
−
1
−
e
Q
z
.
Similarly as above, we getFz−eQz−1/eQz+1 just has zeros with multiplicities at least 3. Furthermore, we have Fz−1−eQz−c just has zeros with multiplicities at least 3. Suppose that 1−eQz−c,0,eQz−1/eQz+1 are distinct from each other. Then, by the second main theorem, we have			(36)2Tr,F≤N¯r,F+N¯r,1F+N¯r,1F−1−eQz−c+N¯r,1F−eQz−1/eQz+1+Sr,F≤13Nr,F+13N¯r,1F+13Nr,1F−1−eQz−c+13Nr,1F−eQz−1/eQz+1+Sr,F≤43Tr,F+Sr,F,
a contradiction. Thus, there are at lest two functions equal in the set(37)1−eQz−c,0,eQz−1eQz+1.
We consider two cases.
Case 1. 1−eQz−c≡0 or eQz−1/eQz+1≡0 and then eQz−1≡0, then Q must be a constant with eQz≡1, so Fz+c−Fz≡Fz, and then Fz+c=2Fz, at last f3z+c=2f3z.
Then, we haveeQ=1. It follows from (33) that the assertion holds.
Case 2. 1−eQz−c=1−e−Qz. That is eQz−c+Qz=1.
Noting thatQ is a polynomial, we deduce that Qz is a constant by the assumption of Case 2. Furthermore, e2Q=1 and eQ=−1. By (33), we get f3z+f3z+c=2. Then, it follows from Theorem 1 that the case cannot occur.
Thus, we finish the proof of Theorem5.Example 1.
Considerfz=2sinz. Then fz+c=2cosz for c=π/2, and Δcfz=fz+c−fz=2sinπ/4−z. Notice that fz, Δcfz share 2,−2, ∞ CM, but fz≠±fz+c.



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*Source: 3205357-2020-02-02.xml*