COMPLETE CLASSIFICATION OF MEROMORPHIC SOLUTIONS OF A FERMAT-TYPE DIFFERENCE EQUATION

Abstract We investigate meromorphic solutions of the difference equation f ( z ) 2 + f ( z + c ) = h ( z ) , c ∈ C ∖ { 0 } , $$ \begin{align*} f(z)^2+f(z+c)=h(z), \quad c\in\mathbb C\setminus\{0\}, \end{align*} $$ where h is a transcendental meromorphic function satisfying N ( r , h ) + N ( r , 1 / h ) = S ( r , h ) . $ N(r,h)+N(r,{1}/{h})=S(r,h). $ upper N left parenthesis r comma h right parenthesis plus upper N left parenthesis r comma 1 divided by h right parenthesis equals upper S left parenthesis r comma h right parenthesis period We prove that every solution of hyper-order less than 1 $1$ 1 is of the form f ( z ) = e a z + b − 1 , h ( z ) = e 2 a z + 2 b , $$ \begin{align*} f(z)=e^{az+b}-1, \quad h(z)=e^{2az+2b}, \end{align*} $$

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Publication Details

Journal
Bulletin of the Australian Mathematical Society
Published
2026-10-30
DOI
https://doi.org/10.1017/s0004972726102202
Primary Topic
Meromorphic and Entire Functions
Type
article
Field-Weighted Citation Impact
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article

COMPLETE CLASSIFICATION OF MEROMORPHIC SOLUTIONS OF A FERMAT-TYPE DIFFERENCE EQUATION

Oleksandr Volkov
Bulletin of the Australian Mathematical Society
Meromorphic and Entire Functions
article

COMPLETE CLASSIFICATION OF MEROMORPHIC SOLUTIONS OF A FERMAT-TYPE DIFFERENCE EQUATION

Oleksandr Volkov
article en

Abstract

Abstract We investigate meromorphic solutions of the difference equation f ( z ) 2 + f ( z + c ) = h ( z ) , c ∈ C ∖ { 0 } , $$ \begin{align*} f(z)^2+f(z+c)=h(z), \quad c\in\mathbb C\setminus\{0\}, \end{align*} $$ where h is a transcendental meromorphic function satisfying N ( r , h ) + N ( r , 1 / h ) = S ( r , h ) . $ N(r,h)+N(r,{1}/{h})=S(r,h). $ upper N left parenthesis r comma h right parenthesis plus upper N left parenthesis r comma 1 divided by h right parenthesis equals upper S left parenthesis r comma h right parenthesis period We prove that every solution of hyper-order less than 1 $1$ 1 is of the form f ( z ) = e a z + b − 1 , h ( z ) = e 2 a z + 2 b , $$ \begin{align*} f(z)=e^{az+b}-1, \quad h(z)=e^{2az+2b}, \end{align*} $$

Bulletin of the Australian Mathematical Society
University of California, Berkeley (US)
Openalex Percentile: Top 7%
Meromorphic and Entire Functions
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