LinkBack URL
About LinkBacksLetbe a function such that
for all prime numbers
and
=f(m)f(n), \ \forall m, n \in \mathbb{N}.)
Prove that for everythere exist
such that
and
Remark: This problem is an easy case of a long standing conjecture in number theory: ifand
is "eventually constant", then
is a perfect square.
Let
Prove that for every
Remark: This problem is an easy case of a long standing conjecture in number theory: if
We have the identity
which shows that it's impossible for the sequenceto be eventually constantly equal to -1. If it is eventually constantly equal to 1, let
be the greatest integer such that
. Then by hypothesis
and then
, contradicting the choice of
On a side note, I believeis called Liouville's function.
You always have nice problems, NonCommLion. Where do you get them from?
Originally Posted by Bruno J.
We have the identity
which shows that it's impossible for the sequencenice!just note that before choosing the greatest integer
is non-empty.
this, of course, is trivial in our case becausebut it becomes absolutely non-trivial if we replace
with an arbitrary non-zero integer
finally the identity you used can be extended. in general we have:((n+1)^2+k)=(n^2+n+k)^2+k.)
You always have nice problems, NonCommLion.thanks! i like my new user name: NCL! it could also stand for Normal CLosure!
Where do you get them from?
here and there!
  |
