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