Entire solutions of f^2+g^2=1
Alan Horwitz
I am interested in all entire solutions f and g to f^2+g^2=1. I remember
seeing this somewhere, but I cannot recall where. In particular, I want
to know if the entire function f+g can have finitely many zeroes.
Paul Vojta
In article <4r9kv3$1a...@hearst.cac.psu.edu>,
See work of Fred Gross (at U of Maryland Baltimore Cty.). Especially,
Bull. AMS vol. 74 and Indian Journal of Math. vol. 23.
--Paul Vojta, vo...@math.berkeley.edu
Robert Israel
In article <4r9kv3$1a...@hearst.cac.psu.edu>, Alan Horwitz <al...@psu.edu> writes:
|> I am interested in all entire solutions f and g to f^2+g^2=1. I remember
|> seeing this somewhere, but I cannot recall where.
to one of my classes, but I don't recall the source. The solutions are
f = cos(h) and g = sin(h) where h is entire. The way to see this is that
(f + i g)(f - i g) = 1 so f + i g is entire and has no zeros, therefore
f + i g = exp(i h) for some h. Then f - i g = exp(-i h), etc.
|> In particular, I want
|> to know if the entire function f+g can have finitely many zeroes.
Since it is cos(h)+sin(h) = sqrt(2) sin(h + pi/4), f+g will have zeros
wherever h takes the values (n-1/4) pi for integers n. By the little
Picard theorem, h can't more than one of these values unless it is constant.
So all nonconstant solutions have infinitely many zeros.
Robert Israel isr...@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Y4
Harold P. Boas
Robert Israel wrote:
>
> In article <4r9kv3$1a...@hearst.cac.psu.edu>, Alan Horwitz <al...@psu.edu> writes:
> |> I am interested in all entire solutions f and g to f^2+g^2=1. I remember
> |> seeing this somewhere, but I cannot recall where.
> I've also seen this before, in fact I recall assigning it as homework
> to one of my classes, but I don't recall the source. The solutions are
Robert B. Burckel gives some history about this problem in his
comprehensive book An Introduction to Classical Complex Analysis,
volume 1 (Academic Press, 1979). In Theorem 12.20, pages 433-435,
he shows that the equation f^n+g^n=1 has no nonconstant entire
solutions when the integer n exceeds 2; when n=2, the solution
is as given by R. Israel in his post.
In the notes on page 458, Burckel attributes the theorem to a 1916
paper of P. Montel, and he gives some further results and references.
P. Vojta, in his post, mentioned work on this subject by Fred Gross.
Some precise references to papers by Gross are:
Further results on Fermat type equations and a characterization
of the Chebyshev polynomials (with C. F. Osgood), J. Math. Anal.
Appl. 121 (1987), no. 2, 317-324; Addendum in the same journal,
134 (1988), no. 1, 254-255.
On the functional equation f^n+g^n=h^n and a new
approach to a certain class of more general functional equations
(with C. F. Osgood), Indian J. Math. 23 (1981), no. 1-3, 17-39.
On the equation f^n+g^n=1. II. Bull. Amer. Math. Soc. 74 (1968),
647-648; Errata, same volume, 767.
On the functional equation f^n+g^n=h^n, Amer. Math. Monthly 73
(1966), 1093-1096.
--
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Harold P. Boas mailto:bo...@math.tamu.edu
World-Wide Web URL: http://www.math.tamu.edu/~Harold.Boas/
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Alan Horwitz
isr...@math.ubc.ca (Robert Israel) wrote:
>
> In article <4r9kv3$1a...@hearst.cac.psu.edu>, Alan Horwitz <al...@psu.edu> writes:
> |> I am interested in all entire solutions f and g to f^2+g^2=1. I remember
> |> seeing this somewhere, but I cannot recall where.
> I've also seen this before, in fact I recall assigning it as homework
> to one of my classes, but I don't recall the source. The solutions are
> (f + i g)(f - i g) = 1 so f + i g is entire and has no zeros, therefore
> f + i g = exp(i h) for some h. Then f - i g = exp(-i h), etc.
>
> |> In particular, I want
> |> to know if the entire function f+g can have finitely many zeroes.
>
> Since it is cos(h)+sin(h) = sqrt(2) sin(h + pi/4), f+g will have zeros
> wherever h takes the values (n-1/4) pi for integers n. By the little
> Picard theorem, h can't more than one of these values unless it is constant.
> So all nonconstant solutions have infinitely many zeros.
>
>
> Robert Israel isr...@math.ubc.ca
> Department of Mathematics (604) 822-3629
> University of British Columbia fax 822-6074
> Vancouver, BC, Canada V6T 1Y4
>
Now for another question about entire functions. Let f(z) be an odd
entire function. Can there be a complex number C such that f(z)-C has no solution-
i.e. can f omit any values ?
Robert Israel
|> Now for another question about entire functions. Let f(z) be an odd
|> entire function. Can there be a complex number C such that f(z)-C has no solution-
|> i.e. can f omit any values ?
No, unless it's identically 0. Use little Picard: if it omits c then it omits -c,
so the only possibility is to omit 0. But f(0)=0 because f is odd.
Bill Dubuque
"Harold P. Boas" <bo...@math.tamu.edu> wrote to sci.math.research on 7/3/96:
:
:Robert Israel wrote:
:>
:> |> I am interested in all entire solutions f and g to f^2+g^2=1.
:> |> I remember seeing this somewhere, but I cannot recall where.
:>
:> I've also seen this before, in fact I recall assigning it as homework
:
:Robert B. Burckel gives some history about this problem in his
:comprehensive book An Introduction to Classical Complex Analysis,
:volume 1 (Academic Press, 1979). In Theorem 12.20, pages 433-435,
:he shows that the equation f^n+g^n=1 has no nonconstant entire
:solutions when the integer n exceeds 2; when n=2, the solution
Note that the rational function case of FLT follows trivially from
Mason's abc theorem, e.g. see Lang's Algebra, 3rd Ed. p. 195 for a
short elementary (high-school level) proof of both. Chebyshev also
gave a proof of FLT for poly's via the theory of integration in
finite terms, e.g. see p. 145 of Shanks' "Solved and Unsolved Problems
in Number Theory", or Ritt's "Integration in Finite Terms", p. 37.
The Chebyshev result is actually employed as a subroutine of Macsyma's
integration algorithm (implemented decades ago by Joel Moses). Via abc
a related result of Dwork is also easily proved: if A,B,C are fixed
poly's then all coprime poly solutions of A*X^a+B*Y^b+C*Z^c = 0
have bounded degrees provided 1/a+1/b+1/c < 1. Other applications
in both number and function fields may be found in Lang's survey [3].
Mason's abc theorem may be viewed as a very special instance of a
Wronskian estimate: in Lang's proof the corresponding Wronskian
identity is c^3*W(a,b,c) = W(W(a,c),W(b,c)), thus if a,b,c are
linearly dependent then so are W(a,c),W(b,c); the sought bounds
follow upon multiplying the latter dependence relation through by
N0 = r(a)*r(b)*r(c), where r(x) = x/gcd(x,x').
More powerful Wronskian estimates with applications toward
diophantine approximation of solutions of LDEs may be found in
the work of the Chudnovsky's [1] and C. Osgood [2]. References
to recent work may be found (as usual) by following MR citations
to these papers in the MathSci database.
I have not seen mention of this Wronskian view of Mason's abc theorem.
Although elementary, it deserves attention since it connects the abc
theorem with the general unified viewpoint of the Wronskian formalism
as proposed by the Chudnovsky's and others.
[1] Chudnovsky, D. V.; Chudnovsky, G. V.
The Wronskian formalism for linear differential equations and Pade
approximations. Adv. in Math. 53 (1984), no. 1, 28--54.
86i:11038 11J91 11J99 34A30 41A21
[2] Osgood, Charles F.
Sometimes effective Thue-Siegel-Roth-Schmidt-Nevanlinna bounds, or better.
J. Number Theory 21 (1985), no. 3, 347--389. 87f:11046 11J61 12H05
[3] Lang, Serge
Old and new conjectured Diophantine inequalities. Bull. Amer. Math. Soc.
(N.S.) 23 (1990), no. 1, 37--75. 90k:11032 11D75 11-02 11D72 11J25
-Bill