Re: d|a[n]
- From: Bill Dubuque <wgd@xxxxxxxxxxxxxxxxxxxx>
- Date: 24 Apr 2009 11:42:43 -0400
achille <achille_hui@xxxxxxxxxxxx> wrote:
On Apr 24, 6:44 pm, Patrick Coilland <pcoill...@xxxxxx> wrote:
KY a écrit :
(1)Prove n^5-5*n^3+4*n is divisible by 120,
for every integer n
(2)Prove that for any natural number n
17^n-12^n-24^n+19^n is divisible by 35.
Must we use
http://mathworld.wolfram.com/Integer-RepresentingPolynomial.html ??
IMHO, the recurrence method metioned by Bill is much simplier to learn
and easier to use than those complicated representing polynomials.
Worth mention is that the classical Polya-Ostrowski results on
integer-valued polynomials have been widely generalized, e.g. see below
------------------------------------------------------------------------------
46#9027 13F20
Cahen, Paul-Jean
Polynomes valeurs entieres. (French)
Canad. J. Math. 24 (1972), 747--754.
------------------------------------------------------------------------------
Let A be a Dedekind domain with field of fractions K . Let A_S denote
the ring of polynomials in K[x] that map A into itself; A_S was studied
earlier by Polya (reference below) and Ostrowsky. THEOREM: There is a sequence
of monic polynomials f_0,f_1,... in K[x] and a sequence of fractional
ideals I_0,I_1,... of A such that f_n are degree n for all n , and
\sum_n k_n f_n in A_S (k_n in K) if and only if k_n in I_n for all n .
Such f_n and I_n are first constructed for valuation rings;
the approximation theorem is applied to obtain the general case. The
fractional ideals I_n are uniquely determined and the author gives several
characterizations of them. A_S is their direct sum (as an A-module) and
hence its projective. The author considers other "decompositions" of A_S
into a sum of fractional ideals of A with respect to sequences of
polynomials of increasing degree that are not necessarily monic. COROLLARY
[G. Polya, J. Reine Angew. Math. 149 (1919), 97-116; Jbuch 47, 163]:
A_S is free on a set of polynomials of increasing degree if and only if the
I_n are principal. COROLLARY: If A is the ring of integers of an algebraic
number field and P in A_S has degree n then n!P in A[x] .
Reviewed by Ron Brown
------------------------------------------------------------------------------
JFM 47.0163.04
Polya, G.
Uber ganzwertige Polynome in algebraischen Zahlkorpen
[J] J. fur Math. 149, 97-116. Published: (1920)
JFM 47.0163.05
Ostrowski, A.
Uber ganzwertige Polynome in algebraischen Zahlkorpen
[J] Ebenda, 117-124. Published: (1920)
http://www.emis.de/cgi-bin/JFM-item?47.0163.05
Google translation:
A polynomial P(x) over a field K is called integer-valued, if P(\xi) is
integral for each integere \xi from K. It is now asked, when in K a
consequence of integer-valued polynomials - a regular basis P_0(x), P_1(x),
...., P_1(x), ... of the degrees of 0,1,...,i,... exists with the characteristic
that each integer-valued polynomial in K can be built up from these polynomials
with integer coefficients from K. Each integer-valued polynomial in K of degree
m can be written in the form 1/m! (\a x^m + \b x^{m-1} +...+\l} where \a,\b,...
,\l are integral. The integral numbers \a in this representation form an ideal
{\\V}_m, and which is necessary and sufficient condition for the existence of
a regular basis that all ideals {\\V}_m belong to the main class. Further
discussion shows that all are {\\V}_m in a quadratic field then and main
ideals only if all prime ideal divisors of the discriminant are it.
far one solves the Verf. the question, which is the
highest power \\p^\psi(\\p) a prime ideal \\p, which ten for all
whole \xi from K the values of an integral polynomial P(x) m degree divide can,
if all coefficients of P(x) are not divisible through \\p. The result results
\psi(\\p) = \sum_{i=1}^m [m/N(\\p^i)]. With the
help of these results then one shows, how a regular basis for the fields K with
the class number of 1 is to be designed \par The question about the conditions
for the existence of a regular basis for integer-valued polynomials, raised
in the first work of G. Polya, in second of A. Ostrowski decided for any
finite field generally. As the necessary and sufficient condition arises:
If one forms the product of all prime ideals of the same degree, which divides
a rational prime number p, then this product for each degree and for each p
must belong to the main class. - for normal fields the existence of the basis
depends only on the discriminant divisors. E.g. exists a regular basis in all
by the disjunction primitive a unit root of prime number power degrees to
developing number fields. The same investigation is accomplished for
integer-valued polynomials in several variables and it arises the same
condition for the existence of one suitably defined regular basis
--Bill Dubuque
.
- References:
- d|a[n]
- From: KY
- Re: d|a[n]
- From: Patrick Coilland
- Re: d|a[n]
- From: achille
- d|a[n]
- Prev by Date: Eliminate
- Next by Date: Re: Some questions on Lattices
- Previous by thread: Re: d|a[n]
- Next by thread: Re: Path on complex plane
- Index(es):
Relevant Pages
|
