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).

Question

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 · Answer

anyone... i m lying dead!!

anonymous · Answer

I think the answer is p(x) = x + 1.

anonymous · Answer

k. i simplify the problem. PROVE THAT p(x)=x+1.

anonymous · Answer

well jprahman r u a bot or a prodigy?

anonymous · Answer

how will u prove it...u are correct :)

anonymous · Answer

The sequence formed by x1 = 1, x2 = p(x1) ... can be rewritten as x(n) = n, therefore x(N) is divisible by N

anonymous · Answer

explanations jprahman?

anonymous · Answer

Well, 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 · Answer

hey how is p(x1)=1+1=2?

anonymous · Answer

isnt that a wrong assumption?

anonymous · Answer

no, 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 · Answer

explain more....

anonymous · Answer

what are u trying to prove by induction?

anonymous · Answer

notice that the nth term is p(p(p(.....(1).....)

anonymous · Answer

Well we see that x(n) = n. Since p(x(n - 1)) = n and x(n) = n then p(x(n-1)) = p(n - 1). Now we have p(n - 1) = n, therefore p(n) = n + 1

anonymous · Answer

how x(n)=n?

anonymous · Answer

Ok, 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(n-1)) 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 · Answer

You 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 · Answer

Divisibility 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.