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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).

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  1. anonymous
    • 5 years ago
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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

  2. anonymous
    • 5 years ago
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    hmmm...are u sure x2 is an integer? since u used p(x2)>x2?

  3. anonymous
    • 5 years ago
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    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?

  4. anonymous
    • 5 years ago
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    dindatc any ideas?

  5. anonymous
    • 5 years ago
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    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)

  6. anonymous
    • 5 years ago
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    if i use p(x2)>x2 , the x2 has to be a positive integer

  7. anonymous
    • 5 years ago
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    yes but tats not give :( any number is divisible by 1. so p(1) being divisible by 1 is not useful..u know...

  8. anonymous
    • 5 years ago
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    yes, sorry i dont know how that linked to find p(x)

  9. anonymous
    • 5 years ago
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    i think we have to find some inference from the given data before jumping to the solution..Its a good one i think

  10. anonymous
    • 5 years ago
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    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

  11. anonymous
    • 5 years ago
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    see that p(x) has integer coefficients so p(1) is indeed an integer. so we have x1,x2...as integers

  12. anonymous
    • 5 years ago
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    every no. is divisible by 1.

  13. anonymous
    • 5 years ago
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    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.

  14. anonymous
    • 5 years ago
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    now we have p(p(p...(1)))....)>....>p(p(1))>p(1)>1. What do we get from here? Any ideas dindatc?

  15. anonymous
    • 5 years ago
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    hm, no idea. maybe that has something to do with infinite sequence

  16. anonymous
    • 5 years ago
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    also how to apply the fact that one of the terms in the sequence is divisible by a positive integer?

  17. anonymous
    • 5 years ago
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    I dont think that infinite sequences will be used here dindatc..Its a precalculus problem!

  18. anonymous
    • 5 years ago
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    soryy, i can't help, i really dont have any ideas

  19. anonymous
    • 5 years ago
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    neither do I dindatc:(

  20. anonymous
    • 5 years ago
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    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?

  21. anonymous
    • 5 years ago
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    and how can that help us find p(x)?

  22. anonymous
    • 5 years ago
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    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...

  23. anonymous
    • 5 years ago
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    <<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..!>>

  24. anonymous
    • 5 years ago
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    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

  25. anonymous
    • 5 years ago
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    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

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

  27. anonymous
    • 5 years ago
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    if you continue : p(p(p(1)))>p(p(1)) ->p(p(2))>p(2) ->p(3)>3 ->p(3)=4

  28. anonymous
    • 5 years ago
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    because p(2)>2 so p(2) is any number greater than 2, isnt it?

  29. anonymous
    • 5 years ago
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    that's the same as when you pick p(1)=2

  30. anonymous
    • 5 years ago
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    right it could be 7 or 100 or 134534

  31. anonymous
    • 5 years ago
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    why p(1)=2 why not 7,or anything>2. That we have to prove..

  32. anonymous
    • 5 years ago
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    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,.......

  33. anonymous
    • 5 years ago
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    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)

  34. anonymous
    • 5 years ago
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    yeah, that's right. i think p(x) = x+1 is right

  35. anonymous
    • 5 years ago
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    yah thats right..cause its given to prove in the problem:) lol. But we have to prove this.

  36. anonymous
    • 5 years ago
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    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

  37. anonymous
    • 5 years ago
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    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

  38. anonymous
    • 5 years ago
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    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)

  39. anonymous
    • 5 years ago
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    ya..but can be like p(1)=3,p(2)=2,p(3)=4,p(4)=5...

  40. anonymous
    • 5 years ago
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    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)

  41. anonymous
    • 5 years ago
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    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.

  42. anonymous
    • 5 years ago
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    p(2) always has a value. but by taking p(1)=3 we cant KNOW it...

  43. anonymous
    • 5 years ago
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    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)

  44. anonymous
    • 5 years ago
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    How do u know p(p(1))=p(2) and how do u know p(3)=p(p(p(1)))?

  45. anonymous
    • 5 years ago
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    from the sequence p(1),p(2),p(3),p(4),...... p(1),p(p(1)),p(p(p(1))),.. CMIIW

  46. anonymous
    • 5 years ago
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    i gotta go, sorry can't continue the discussion. hope you find the answer soon :)

  47. anonymous
    • 5 years ago
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    k..thnx:) bye.

  48. anonymous
    • 5 years ago
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    So I am all alone now... i think :)

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