Question

It is known of a polynomial over Z that p(n)>n for all positive integer n. Consider x1=1,x2=p(x1),...
We know that, for any positive integer N, there exists a term of the sequence divisible by N.Find p(x).

Answer 1

i dont know how to find p(x), but i think if p(n)>n for all positive integer n, then p(x)>x for all positive integer x.
x1=1 -> p(x1)>x1 -> p(1)>1
x2=p(x1)-> p(p(x1))>p(x1) -> p(p(1))>p(1) <- i'm stuck here

Answer 2

hmmm...are u sure x2 is an integer? since u used p(x2)>x2?

Answer 3

the sequence 1,p(1),p(p(1)),p(p(p(1))),...has a term divisible by any positive integer. So there are terms divisible by 2,3,4,5,6,...How do we use this fact?

Answer 4

dindatc any ideas?

Answer 5

for any positive integer N, there exists a term of the sequence divisible by N
we know hat p(n)>n
for x1->p(1)>1
then p(1) must be divisible by 1.
if p(1)>1 , that means we can choose any number that is bigger than 1 as the value of p(1)

Answer 6

if i use p(x2)>x2 , the x2 has to be a positive integer

Answer 7

yes but tats not give :(
any number is divisible by 1. so p(1) being divisible by 1 is not useful..u know...

Answer 8

yes, sorry i dont know how that linked to find p(x)

Answer 9

i think we have to find some inference from the given data before jumping to the solution..Its a good one i think

Answer 10

the sequence 1,p(1),p(p(1)),p(p(p(1))),.. is right, but it isnt divisible by 2,3,4,5,6,..., it's divisible by 1,2,3,4,....
because the x1 starts from 1

Answer 11

see that p(x) has integer coefficients so p(1) is indeed an integer.
so we have x1,x2...as integers

Answer 12

every no. is divisible by 1.

Answer 13

Well we got that the sequence consists of integers. But we dont know whether the sequence contains only positive integers!
But observe that since 1 is a positive integer then p(1)>1.
So the sum of the coefficients of the polynomial is >1
That means that p(1)>0. So this implies p(p(1))>0. And therefore
the sequence has all positive integers.
so now we can use p(n)>n to the sequence.

Answer 14

now we have p(p(p...(1)))....)>....>p(p(1))>p(1)>1.
What do we get from here? Any ideas dindatc?

Answer 15

hm, no idea.
maybe that has something to do with infinite sequence

Answer 16

also how to apply the fact that one of the terms in the sequence is divisible by a positive integer?

Answer 17

I dont think that infinite sequences will be used here dindatc..Its a precalculus problem!

Answer 18

soryy, i can't help, i really dont have any ideas

Answer 19

neither do I dindatc:(

Answer 20

the main problem now is: from the sequence
p(p(p...(1)))....)>....>p(p(1))>p(1)>1 how can we use that there exists terms that are divisible by any positive integer?

Answer 21

and how can that help us find p(x)?

Answer 22

but wait ..i said x2 is a positive integer. so p(1) is an integer (see few posts befor the last, its proven).
Clearly p(1)>1 so p(1)=2,3,4,..
Now p(1) is divisible by atleast one of 2,3,4,5,6,...(recall the given condition!)
dindatc, i want to find p(1) in aim of discovering the sum of the coefficients...

Answer 23

<<the problem was to prove that p(x)=x+1 , which i told u to find in the question. So if we find p(1)=2 then we are done..!>>

Answer 24

but wait...if p(1)=2 it does not necessarily imply that p(x)=x+1.We also have to show along with p(1)=2 that degree of p(x)=1

Answer 25

if you choose p(1)=2, then p(2)=3, p(3)=4.....
as p(1)>1 ->p(1)=2
p(p(1))>p(1) -> p(2)>2 ->p(2)=3
that proves p(x)=x+1

Answer 26

ooo...thats too faulty...but good try nonetheless..
if p(1)=2 how do we know p(2)=3 or not?

Answer 27

if you continue :
p(p(p(1)))>p(p(1)) ->p(p(2))>p(2) ->p(3)>3 ->p(3)=4

Answer 28

because p(2)>2
so p(2) is any number greater than 2, isnt it?

Answer 29

that's the same as when you pick p(1)=2

Answer 30

right it could be 7 or 100 or 134534

Answer 31

why p(1)=2 why not 7,or anything>2. That we have to prove..

Answer 32

let's try another value of p(1)
assume p(1)=7
p(p(1))>p(1) ->p(7)>7 -> p(7) can be 8,9,10,.......

Answer 33

ya..good thinking! we have p(1)=7 and p(7)=anything>7
so we dont know anything about p(2) or p(3) or....p(6)

Answer 34

yeah, that's right. i think p(x) = x+1 is right

Answer 35

yah thats right..cause its given to prove in the problem:) lol.
But we have to prove this.

Answer 36

p(1)>1 ->p(1)=2
p(p(1))>p(1) -> p(2)>2 ->p(2)=3
p(p(p(1)))>p(p(1)) ->p(p(2))>p(2) ->p(3)>3 ->p(3)=4
the sequence is 2,3,4,........
so p(x)= x+1
i think that's enough to prove it

Answer 37

dindatc, p(1)>1 doesnt imply that p(1)=2! p(1) could well have been 3 also. You need to prove that p(1)=2

Answer 38

no, if you pick another number for the value of p(1), you won't have a complete sequence
just like when we pick p(1)=7
we can't get p(2),p(3),p(4),p(5), and p(6)
if you choose p(1)=3, you won't get the value of p(2)

Answer 39

ya..but can be like p(1)=3,p(2)=2,p(3)=4,p(4)=5...

Answer 40

the second term is p(x2) , x2=p(x1), and x1=1
so p(p1)>p(1)
if p(1) = 3 then :
p(3)>3 , you jumped from p(1) tp p(3), you can't get the value of p(2)

Answer 41

u dont know if p(2)>p(1) so p(2) can be less than p(1)...
if u took p(1)=3 then u get p(3)>3.

Answer 42

p(2) always has a value. but by taking p(1)=3 we cant KNOW it...

Answer 43

the sequence you post above : p(1),p(p(1)),p(p(p(1))),..
so i think so i think p(3) is greater than p(2)
because p(2) is p(p(1))
and p(3) is p(p(p(1))) , p(p(p(1)))>p(p((1))
this proves that p(3)>p(2)

Answer 44

How do u know p(p(1))=p(2) and how do u know p(3)=p(p(p(1)))?

Answer 45

from the sequence
p(1),p(2),p(3),p(4),......
p(1),p(p(1)),p(p(p(1))),..
CMIIW

Answer 46

i gotta go, sorry can't continue the discussion. hope you find the answer soon :)

Answer 47

k..thnx:) bye.

Answer 48

So I am all alone now... i think :)