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From: Martin M. + I am right:if I am not tell me:or I am right on 27 May 2010 12:20
On 27 May, 04:59, Deep <deepk...(a)yahoo.com> wrote:
> On May 26, 10:46 pm, The Pumpster <pumpledumplek...(a)gmail.com> wrote:
>
>
>
>
>
> > On May 26, 6:48 pm, James Waldby <n...(a)no.no> wrote:
>
> > > On Wed, 26 May 2010 17:31:14 -0700, Deep wrote:
> > > > Consider the following equation under the given conditions.
> > > > m = (f^k + e^k)/(2k)     (1)
>
> > > > m, f, e are integers each > 1, k is a prime > 3; f, e are odd and
> > > > coprimes.
>
> > > Let (2) stand for those conditions on e,f,k,m.
>
> > > > Conjecture:  m can assume infinitely many integer values for infinite
> > > > choices of e and f.
>
> > > ...
>
> > > If your conjecture is that "The set S is infinite", where
> > > S = {(e,f): there are infinitely many k, m such that e,f,k,m
> > > satisfy (1) and (2) }, I think it is false, as follows.
>
> > > Suppose e,f,k,m satisfy (1) and (2); then 2*k*m = e^k + f^k.
>
> > > By flt*, e^k == e mod k and f^k == f mod k.
>
> > > When k > e + f, we have 0 =/= e+f == e^k + f^k mod k.
>
> > > But 2*k*m == 0 mod k, a contradiction when k > e + f.
>
> > > Hence, for any given e,f there can be only finitely many
> > > k, m such that e,f,k,m satisfy (1) and (2); so S is empty.
>
> > > *<http://en.wikipedia.org/wiki/Fermat%27s_little_theorem>
> > > --
> > > jiw
>
> > So all one needs is to take e = -f mod k, to have k divide e+f.
>
> > de P- Hide quoted text -
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> > - Show quoted text -
>
> *****
> One simple numerical example will be appreciated.
>
> ******- Hide quoted text -
>
> - Show quoted text -

Jump to I. The Period of F(mod m)‎: Thus lcm[k(piei)] | k(m). Let L =
lcm[k(piei)]. Since k(piei) | L for each i, the sequence F(mod
piei) ...On the Ramanujan-Petersson conjecture for modular forms of
half ...by W Kohnen
(f,e~..,> = Ckm-k+l/2a(m) (VfeS~+l,2), .... (Ramanujan-Petersson
conjecture) holds for all f = ~, a(n)q"ES~+ 1/2 if and only if the
estimate ...Collatz conjecture [edit] m-cycles. The proof of the
conjecture can indirectly be done by proving it is easy to check f (k)
< k , so f (k) ∈ E; hence k ∈ E. ...
A conjecture on the number of conjugacy classes in a <Emphasis ...
Now, if M < G, again by induction, we have that k(M) < INIp, ... l-I
k(Ev/Z) = k(E/Z). But E C_ F and in fact E = F since Ep contains
_pp. ...
Noncommutative-Nilpotent Matrices and the Jacobian Conjecture
polynomial map of k" into itself. The Jacobian conjecture says F is
an ... nilpotent if and only if there is a T E GL, (k) , such 7-l M
( X ) T is [PDF] On the Stanley-Wilf conjecture for the number of
permutations
Conjecture 1. ($100.00) For all σ ∈ Sk and n ≥ 1, F(n, σ) ≤ (k −
1)2n. ... iff σ(a) < σ(b). Conjecture 2. As k → ∞, m(k) ∼ (k/e)2 . ...
The Fibonacci Sequence Modulo M, Jump to I. The Period of F(mod m)‎:
Thus lcm[k(piei)] | k(m). Let L = lcm[k(piei)]. Since k(piei) | L for
each i, the sequence F(mod piei) ...
ON A CONJECTURE OF CHVATAL ON m-INTERSECTING HYPERGRAPHS system S^t,
k, v)), then the conjecture of Chvatal is true for the triple ... (e)
m = 2,h and n = h(4h + l)/5. (h $* 5). (f) m = 2,handn = h(((k- l)h
+ ...
322_p 355..358 The vector p e fk, 2kgE…P†p…e† ˆ 2k if and only if e e
M is ... By Lemma 3.2, if Conjecture 3.1 is true for k even, then
Conjecture 3.1 is true ...ON THE CASAS-ALVERO CONJECTURE 1. The
problem Eduardo Casas-Alvero Conjecture 1.1. Let K be a field of
characteristic 0 and let f ∈ K[x] be a monic ... so f = xn + s2. But
this is a pe-th power in K[x]. ... is non-negative, and let M be the
maximal ideal of O. Hence O/M is a field of ...Transcendence
conjectures about periods of modular forms and ...The K-vector spaces
M~ and Mfk(K) define K-rational structures on M2k , i.e. ... e,g. the
periods of a normalized primitive Hecke eigenform of weight 2


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