- anonymous

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).

- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

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?

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## More answers

- anonymous

dindatc any ideas?

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

every no. is divisible by 1.

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

neither do I dindatc:(

- anonymous

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?

- anonymous

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

- anonymous

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...

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

right it could be 7 or 100 or 134534

- anonymous

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

- anonymous

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,.......

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

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)

- anonymous

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.

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

k..thnx:) bye.

- anonymous

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

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