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anonymous
 5 years ago
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).
anonymous
 5 years ago
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).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm...are u sure x2 is an integer? since u used p(x2)>x2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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
 5 years ago
Best ResponseYou've already chosen the best response.0if i use p(x2)>x2 , the x2 has to be a positive integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but tats not give :( any number is divisible by 1. so p(1) being divisible by 1 is not useful..u know...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, sorry i dont know how that linked to find p(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think we have to find some inference from the given data before jumping to the solution..Its a good one i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0see that p(x) has integer coefficients so p(1) is indeed an integer. so we have x1,x2...as integers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0every no. is divisible by 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0now we have p(p(p...(1)))....)>....>p(p(1))>p(1)>1. What do we get from here? Any ideas dindatc?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hm, no idea. maybe that has something to do with infinite sequence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also how to apply the fact that one of the terms in the sequence is divisible by a positive integer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont think that infinite sequences will be used here dindatc..Its a precalculus problem!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0soryy, i can't help, i really dont have any ideas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0neither do I dindatc:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0and how can that help us find p(x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0<<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..!>>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0ooo...thats too faulty...but good try nonetheless.. if p(1)=2 how do we know p(2)=3 or not?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you continue : p(p(p(1)))>p(p(1)) >p(p(2))>p(2) >p(3)>3 >p(3)=4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because p(2)>2 so p(2) is any number greater than 2, isnt it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's the same as when you pick p(1)=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right it could be 7 or 100 or 134534

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why p(1)=2 why not 7,or anything>2. That we have to prove..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let'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
 5 years ago
Best ResponseYou've already chosen the best response.0ya..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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that's right. i think p(x) = x+1 is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah thats right..cause its given to prove in the problem:) lol. But we have to prove this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0p(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
 5 years ago
Best ResponseYou've already chosen the best response.0dindatc, 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
 5 years ago
Best ResponseYou've already chosen the best response.0no, 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
 5 years ago
Best ResponseYou've already chosen the best response.0ya..but can be like p(1)=3,p(2)=2,p(3)=4,p(4)=5...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0u 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
 5 years ago
Best ResponseYou've already chosen the best response.0p(2) always has a value. but by taking p(1)=3 we cant KNOW it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0How do u know p(p(1))=p(2) and how do u know p(3)=p(p(p(1)))?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from the sequence p(1),p(2),p(3),p(4),...... p(1),p(p(1)),p(p(p(1))),.. CMIIW

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i gotta go, sorry can't continue the discussion. hope you find the answer soon :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I am all alone now... i think :)
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