Question of sequence

Let me give an outline of a proof. The details are not difficult to fill in.

If f(0)=0, then a1=0, and we are done. So suppose f(0)≠0. Let b be the integer root of f(x) with the largest absolute value. (If no such b exists, then obviously a_m cannot be 0 for all m>0.) Then we can write

f(x) = (x-b)g(x). Note that g(x) has integer coeff.

Note that g(0)≠0. Let g(0)=k. If |k|>1. Then we can show by induction that a_m is a nonzero multiple of kb for all m>0, and in particular a_m≠0 for all m>0.

So consider the case |k|=1. If k=-1, then we have a1=b, a2=0, and we are done.

So it remains the case k=1. If g(-b)=0, then we have a1=-b, a2=0, and we are done. So suppose g(-b)≠0. We want to show that a_m≠0 for all m>0. In fact, we can show by induction that

a_m = rb for some integer r not equal to 0 or 1.

It is true for m=1 because a1=-b. For the induction step, if a_m = rb where r =<-1 or r => 3, it is easy to show that a_(m+1) also satisfy the statement. (details skipped here.) So it remains the case r=2. Note that f(2b)≠0, so all we need to do is to prove that f(2b)≠b.

If a_n=2b for some n, let m be the smallest integer such that a_m=2b. By the above discussion, a_(m-1)=rb where are cannot be =<-1 or =>3. Also r is not 0, 1 or 2. Hence we must have a_(m-1)=-b. (ie, m=2)

Assume on the contrary, we have f(2b)=b. Then

(2b-b)g(2b)=b.

so g(2b)=1. But recall that g(0)=1, so we can write g(x)=xh(x)+1. We have (2b)h(2b)=0, so h(2b)=0. Hence h(x)=(x-2b)q(x), where q(x) has integer coeff.

On the other hand, because f(-b)=2b, we have (-b-b)g(-b)=2b, so

g(-b)=-1.

This implies

(-b)(-b-2b)q(2b)+1=-1

3b^2 q(2b)=-2.

LHS is a multiple of 3 but RHS is not, hence this is a contradiction. QED

2009-03-11 16:30:39 補充：

Ivan: 出處？我怎麼知道，應該問發問者啊！

2009-03-11 18:05:34 補充：

致另一位候選者(因已進入投票，看不到是誰):

你的做法和我的基本上一樣，只是用的文字表達不同己而。

g(0)=1的情況是比較複雜，所以我的答案後面一半的篇幅都是處理這個情況。

myisland8132 :

級是一定要升的，累積到一定的答題數就會自動升級，不想升也要升。現在未升是因為沒有這麼多時間答題目。不過升不升級我是不介意啦。

我知道為何你們不升級啦﹐一定是覺得大學級個Label後生些和smart些

借個位問問題：

Let

a0 = 0

a1 = f(0) = c

a2 = f(c)

...

a_(n+1) = f(an)

then an = hn(c) = is a polynomial of c with integer coefficients for n >= 2 and c divides an for n >= 0. ... (*)

If a1=c=0, then we are done. Otherwise, suppose a1≠0.

Since am = f( a_(m-1) ) = hm(c) = 0, we have

f(x) = (x –a_(m-1) ) g(x) = (x –h_(m-1)(c) ) g(x)

where g is a polynomial with integer coefficients.

This implies

c = f(0) = -h_(m-1)(c) g(0)

Comparing coefficients, we have:

Case 1: g(0) = 0 implies a1=f(0)=0

Case 2: h_(m-1)(c) = c and g(0) = -1

Case 3: h_(m-1)(c) = -c and g(0) = 1

For Case 1 & 2, we are done. We need to check Case 3.

我處理不了Case 3。但在繼續之前，想請教各位高手，上面咁做有無問題？

2009-03-08 17:24:06 補充：

都未討論完，點解變投票？

max ideal space: 請問出處?

answer your first question:

an+1 = f(an) = 1+n

an = 1 + (n-1)

a0 = 1 + (0-1)= 0