AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be an odd integer greater than 1 and let 
be integers. For each permutation 
of 
, define 
. Prove that there exist permutations 
of such that 
is a divisor of 
.
be an integer. Let 
and 
be polynomials with real coefficients such that the points 
in 
are the vertices of a regular 
-gon in counterclockwise order. Prove that at least one of and has degree greater than or equal to 
and 
be two infinite sequences of integers. Prove that there exists an infinite sequence of integers 
such that for any positive integer 
, the following holds:![\[
\sum_{k|n} k \cdot z_k^{\frac{n}{k}} = \left( \sum_{k|n} k \cdot x_k^{\frac{n}{k}} \right) \cdot \left( \sum_{k|n} k \cdot y_k^{\frac{n}{k}} \right).
\]](http://latex.artofproblemsolving.com/2/5/9/259ee32c89befe7a3ee2797e0d5b1df1c3066cf6.png)
such that the relation ![\[|z^{2n}+z^nw^n+w^{2n} | = 2^{2n}+2^n+1 \]](http://latex.artofproblemsolving.com/1/2/2/1229a609403904449c447225094b3d97e063efd4.png)
holds for all positive integers .
such that 
for reals 
.
, let 
denote the number of positive divisors of . Determine whether there exists a positive integer triple 
such that there are exactly 
positive integers 
not greater than 
that satisfies the following: the equation![\[ \tau(x) = \tau(y) = \tau(z) = \tau(ax + by + cz) = K \]](http://latex.artofproblemsolving.com/a/8/6/a86ab0dd1aacf681d5d6761a4da73face1647f48.png)
holds for some positive integers 
.
has an incenter 
Suppose 
. Let 
be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points 
and 
be such that is the midpoint of 
and is the midpoint of 
. Point 
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
. Prove that 
.
children sit in a circle facing inward with 
, and each child starts with an arbitrary even number of pencils. Each minute, each child simultaneously passes exactly half of all of their pencils to the child to their right. Then, all children that have an odd number of pencils receive one more pencil.
checkerboard satisfy the following conditions:
checkers have been placed on the board.
, 
is the Gergon point and the incircle 
touch 
, 
, 
at 
, 
, 
respectively.
, 
, 
be 
, 
, 
respectively. in 
and then we get the circle 
in 
and then the circle 
in 
and then the circle 
, , are coaxial.
, the following 
statements are equivalent:
is differentiable, with continuous first derivative.
and for any two sequences 
, convergent to 
, such that 
for any positive integer , the sequence 
is convergent.
![\[\pi^{-s/2} \Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-(1-s)/2} \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s).\]](http://latex.artofproblemsolving.com/d/9/2/d923f607e8751b6aaa1275f8469233e5c6cedc94.png)
If so, I don't understand what you mean by "finding the zeroes of that equation".
and the Gamma Function (which we can assume to be well-understood, in particular regardings its zeroes and poles), the functional equation relates the values of 
and 
.
at 
gives us a zero at 
(unless we are at a pole of the Gamma Function which would cancel the zero, this happens exactly at the even negative integers, giving the so-called "trivial zeroes"). possess some structure: For a zero, also is a zero.
, for a zero, also 
is a zero.
in several ways.
but of course it does not tell us at all whether the zeroes are on that line or not.
(which I assume is a manipulation of the equation in your post). The trivial zeroes are completely obvious in the 
part of the equation. But I now I understand the difficult part about proving the zeroes lie on one line is the part such that 
.
".
, the equation is true, but I don't see why you would study exactly this equation.
help in finding when the equation is equal to zero? Intuitively, I would expect that part is the most nontrivial as it's almost like a recursive function
to and vice versa, but applying it twice really does not give you anything interesting, but a tautology! instead of , this is just a change of variables. is well-understood. For instance, when 
we have the Euler product from which one can read off that there are no zeroes. One can then use the functional equation to "translate" this to the region 
where there will only be the trivial zeroes.
. Again you could translate from 
to 
and vice versa. Indeed, if you can prove that one of them does not contain any zeroes, then the other won't as well. The point is that we understand neither of the two regions and neither of them is a priori simpler than the other!
.
is not sufficient. Computers have been able to detect over a quadrillion non-trivial zeros, all of which have a real component of 
. Just one counterexample would disprove the hypothesis, but until we can mathematically prove that all of the non-trivial zeros have this property, the hypothesis will remain hypothetical.
to and vice versa, but applying it twice really does not give you anything interesting, but a tautology! instead of , this is just a change of variables. is well-understood. For instance, when we have the Euler product from which one can read off that there are no zeroes. One can then use the functional equation to "translate" this to the region where there will only be the trivial zeroes.. Again you could translate from to and vice versa. Indeed, if you can prove that one of them does not contain any zeroes, then the other won't as well. The point is that we understand neither of the two regions and neither of them is a priori simpler than the other!. is not sufficient. Computers have been able to detect over a quadrillion non-trivial zeros, all of which have a real component of . Just one counterexample would disprove the hypothesis, but until we can mathematically prove that all of the non-trivial zeros have this property, the hypothesis will remain hypothetical.
, 
, etc.) equal to zero and find the values of such that it is true. But it seems like the creates a symmetry which makes it more difficult to solve. Is my understanding correct?
. It's just like how computers calculate limits. So they are fund randomly, not from actual calculation. The zeta function is much more irregular than a polynomial like 
. We currently do not have a way of calculating zeros of zeta directly. They are just detected
In other words, does there exist a possibility for a different equation to extend the domain of the Riemann Zeta function?
here. then 
. No more, no less. The other factors (the power of and the sine) are completely irrelevant for that purpose.
be a set of complex numbers. Then 
is an accumulation point of if for every 
, I can find 
with 
such that 
.
has an accumulation point, namely . (The definition of accumulation point generalizes to arbitrary metric spaces, which you'll encounter very often in real analysis.)
be an open region. Then 
is holomorphic if it is "differentiable"; that is the limit 
exists for all 
.
can only be differentiated once. This shows that complex analysis is super lit.
be an open region, be holomorphic. Suppose the zeroes of (i.e. the set 
) has an accumulation point. Then is the zero function.
is holomorphic. is zero everywhere in the unit disk 
, then has to be zero everywhere else because 
has an accumulation point (in fact any point in the closure 
is an accumulation point).
for all 
, then is zero everywhere else too. be an open region, 
be holomorphic. Suppose the set for which 
has an accumulation point. Then 
everywhere.
. 
be open regions with 
. Let 
and 
be holomorphic. We say that 
is the analytic continuation of 
if and agree on 
, i.e. 
. is non-empty, and by virtue of being open (look this up if you're unfamiliar) it definitely has an accumulation point. So if 
are both continuations of to 
, then surely must agree on . By the corollary it follows that and 
agree everywhere, i.e. they're the same analytic continuation. 
Where does this sum actually converge? Write 
. Then if 
then:
Any I think by sending 
the RHS converges via an integral test or something. Then absolute convergence implies convergence, so it follows that converges (and is thus well-defined) for all with real part greater than 
.
and stuff, how could they possibly do that? The short version is that using black magic, you can prove that
and that using additional black magic you can show that 
extends everywhere (this proof is actually pretty cool) and 
extends everywhere except at 
or something, idk I don't remember the proof and this is above my paygrade of zero dollars an hour. Anyways what's important is that an analytic continuation of exists over 
. And by the previous theorem, it's the only possible analytic continuation!
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
