2012
DOI: 10.1017/s1446788712000237
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ON THE EQUATION f(g(x))=f(x)hm(x) FOR COMPOSITE POLYNOMIALS

Abstract: In this paper we solve the equation f (g(x)) = f (x)h m (x) where f (x), g(x) and h(x) are unknown polynomials with coefficients in an arbitrary field K, f (x) is nonconstant and separable, deg g ≥ 2, the polynomial g(x) has nonzero derivative g (x) 0 in K[x] and the integer m ≥ 2 is not divisible by the characteristic of the field K. We prove that this equation has no solutions if deg f ≥ 3. If deg f = 2, we prove that m = 2 and give all solutions explicitly in terms of Chebyshev polynomials. The Diophantine … Show more

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Cited by 2 publications
(3 citation statements)
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References 13 publications
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“…Apart from these results, Conjecture 2 is open. We note that Ganguli and Jankauskas [6] proved related results about sign changes of quadratic polynomials taken at rational values.…”
mentioning
confidence: 74%
See 1 more Smart Citation
“…Apart from these results, Conjecture 2 is open. We note that Ganguli and Jankauskas [6] proved related results about sign changes of quadratic polynomials taken at rational values.…”
mentioning
confidence: 74%
“…What is of utmost importance for us is that the denominators of X, Y, Z are the same, and assuming from this point on that p, q are integers, the numerators of X, Y, Z form an integral solution to (6). (This can also be checked directly.)…”
Section: Proofsmentioning
confidence: 99%
“…It appears to be difficult to classify all such G 0 , G 1 , A 0 , A 1 . In regard to this problem, we mention [10], where the authors solved the equation g(x) f (x) n = g(h(x)) where f , g, h are unknown nonconstant complex polynomials, n > 1, deg h ≥ 2 and g is separable. We also mention [4], where the authors completely classified binomials which have a non-trivial factor which is a composition of two polynomials of degree > 1.…”
Section: Kresomentioning
confidence: 99%