anonymous
  • 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).
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • 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
  • anonymous
hmmm...are u sure x2 is an integer? since u used p(x2)>x2?
anonymous
  • 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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anonymous
  • anonymous
dindatc any ideas?
anonymous
  • 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
  • anonymous
if i use p(x2)>x2 , the x2 has to be a positive integer
anonymous
  • 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
  • anonymous
yes, sorry i dont know how that linked to find p(x)
anonymous
  • 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
  • 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
  • anonymous
see that p(x) has integer coefficients so p(1) is indeed an integer. so we have x1,x2...as integers
anonymous
  • anonymous
every no. is divisible by 1.
anonymous
  • 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
  • anonymous
now we have p(p(p...(1)))....)>....>p(p(1))>p(1)>1. What do we get from here? Any ideas dindatc?
anonymous
  • anonymous
hm, no idea. maybe that has something to do with infinite sequence
anonymous
  • anonymous
also how to apply the fact that one of the terms in the sequence is divisible by a positive integer?
anonymous
  • anonymous
I dont think that infinite sequences will be used here dindatc..Its a precalculus problem!
anonymous
  • anonymous
soryy, i can't help, i really dont have any ideas
anonymous
  • anonymous
neither do I dindatc:(
anonymous
  • 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
  • anonymous
and how can that help us find p(x)?
anonymous
  • 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
  • anonymous
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anonymous
  • 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
  • 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
  • anonymous
ooo...thats too faulty...but good try nonetheless.. if p(1)=2 how do we know p(2)=3 or not?
anonymous
  • anonymous
if you continue : p(p(p(1)))>p(p(1)) ->p(p(2))>p(2) ->p(3)>3 ->p(3)=4
anonymous
  • anonymous
because p(2)>2 so p(2) is any number greater than 2, isnt it?
anonymous
  • anonymous
that's the same as when you pick p(1)=2
anonymous
  • anonymous
right it could be 7 or 100 or 134534
anonymous
  • anonymous
why p(1)=2 why not 7,or anything>2. That we have to prove..
anonymous
  • 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
  • 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
  • anonymous
yeah, that's right. i think p(x) = x+1 is right
anonymous
  • anonymous
yah thats right..cause its given to prove in the problem:) lol. But we have to prove this.
anonymous
  • 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
  • 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
  • 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
  • anonymous
ya..but can be like p(1)=3,p(2)=2,p(3)=4,p(4)=5...
anonymous
  • 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
  • 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
  • anonymous
p(2) always has a value. but by taking p(1)=3 we cant KNOW it...
anonymous
  • 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
  • anonymous
How do u know p(p(1))=p(2) and how do u know p(3)=p(p(p(1)))?
anonymous
  • anonymous
from the sequence p(1),p(2),p(3),p(4),...... p(1),p(p(1)),p(p(p(1))),.. CMIIW
anonymous
  • anonymous
i gotta go, sorry can't continue the discussion. hope you find the answer soon :)
anonymous
  • anonymous
k..thnx:) bye.
anonymous
  • anonymous
So I am all alone now... i think :)

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