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anyone... i m lying dead!!
I think the answer is p(x) = x + 1.
k. i simplify the problem. PROVE THAT p(x)=x+1.
well jprahman r u a bot or a prodigy?
how will u prove it...u are correct :)
The sequence formed by x1 = 1, x2 = p(x1) ... can be rewritten as x(n) = n, therefore x(N) is divisible by N
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.
hey how is p(x1)=1+1=2?
isnt that a wrong assumption?
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.
what are u trying to prove by induction?
notice that the nth term is p(p(p(.....(1).....)
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
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.
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?
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.