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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.0anyone... i m lying dead!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the answer is p(x) = x + 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k. i simplify the problem. PROVE THAT p(x)=x+1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well jprahman r u a bot or a prodigy?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how will u prove it...u are correct :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The sequence formed by x1 = 1, x2 = p(x1) ... can be rewritten as x(n) = n, therefore x(N) is divisible by N

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, if x1 = 1 and x2 = p(x1) = 1 + 1 = 2, then x3 = p(x2) = 2 + 1 = 3. Using mathematical induction we find that the base case is x(1) = 1, and inductively x(n) = p(x(n  1)) = x(n  1) + 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey how is p(x1)=1+1=2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt that a wrong assumption?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, x1 is equal to 1, the base case and first term of the sequence. Subsequent terms are found from the inductive step, for instance x(2) = p(x(2  1) = x(1) + 1 = 2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what are u trying to prove by induction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0notice that the nth term is p(p(p(.....(1).....)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well we see that x(n) = n. Since p(x(n  1)) = n and x(n) = n then p(x(n1)) = p(n  1). Now we have p(n  1) = n, therefore p(n) = n + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, lets start over. In order for there to exist a term T in the sequence that is divisible be any given positive integer then the sequence must be the set of positive integers, or a superset of the positive integers. Because the sequence is defined as x(n) = p(x(n1)) and x(1) = 1 the only possible polynomial over Z such that p(n) > n that will also generate the set of positive integers when applied according to the sequence definition is p(x) = x + 1.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0You need to rewrite your problem Saubhik. It is not so clear. Is this true for every sequence? some sequence? what is that divisibility property?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Divisibility presumably refers to even divisibility. I.e. For every N in the set of positive integers a term T of the sequence exists such that T mod N = 0. The sequence refers to the sequence x(n) generated according to the rule x(n) = p(x(n  1)) where p(n) is the unknown function.
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