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 the positive number satisfied 
Find the minimum of 
, we have 
When does the equality hold?
. If 
, then prove the following: 
people be arranged in a circle, numbered clockwise from 
to 
. These people are eliminated one by one in order, until only one person remains. The elimination follows this rule: among the remaining people, start counting clockwise from the person with the smallest number, and eliminate the -th person in that count. Then, among the remaining people, start counting again from the person with the smallest number and eliminate the -th person. Repeat this process until only one person remains. Let 
denote the number of the last remaining person.
, people are eliminated in the following order: 
. Thus, 
. It is known that 
under certain conditions. Prove that the necessary and sufficient condition for this is that both 
and 
are prime numbers.
be the unit circle in the complex plane. Let 
be the map given by 
. We define 
and 
for 
. The smallest positive integer such that 
is called the period of 
. Determine the total number of points in 
of period 
.
)
and 
and 
+1 w
consecutive even numbers, such that the sum of their squares divides the square of their product.
. Solve the equation: 
be a triangle and let 
be its circumcenter and 
its incenter.
be the radical center of its three mixtilinears and let 
be the isogonal conjugate of .
be the Gergonne point of the triangle .
is parallel with line 
.
be a positive integer. Prove that there exist three permutations 
, 
, and 
of 
such that ![\[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]](http://latex.artofproblemsolving.com/9/8/0/98097fb721722b9103ef65d72322956877e262cb.png)
for every 
.
, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers (
is allowed) such that 
. Find the minimal .
and a point , let 
be the cevian triangle of with respect to . Let 
be the orthocenter of , and denote the isotomic conjugate of 
with respect to by 
, respectively. Let the centroid of be 
, and denote the isogonal conjugate of with respect to by 
. Prove that
or in brief
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
be a primitive 
th root of unity, and observe that for 
, 
. Hence 
, and 
are roots of the quadratic equation 
. A quadratic null at 3 different places must be null everywhere, implying 
, so we're done.
![$T\colon\mathbb R[x]\to\mathbb R^5$](http://latex.artofproblemsolving.com/6/8/8/68855b81fbc161a6f3a1e961dad6f2bb9c1bc2b6.png)
be the function given by![\[ \sum a_kx^k\mapsto\left(\sum_{5\mid k}a_k,\sum_{5\mid k-1}a_k,\ldots,\sum_{5\mid k-4}a_k\right). \]](http://latex.artofproblemsolving.com/5/6/1/561dc24cbb1a22a0740d2ca5cbdea919e9b211a7.png)
It is clear that![\[ T(P(x^5)+xQ(x^5)+x^2R(x^5))=(P(1),Q(1),R(1),0,0). \]](http://latex.artofproblemsolving.com/d/1/9/d191251d5402e6fd85d892a7ece504357b262949.png)
It is also easy to see that![\[ T((x^4+x^3+x^2+x+1)S(x))=(S(1),S(1),S(1),S(1),S(1)). \]](http://latex.artofproblemsolving.com/5/6/8/56868c5e37649b8b4a7b689600853baa00032acc.png)
Since 
, these two must be equal, giving![\[ P(1)=Q(1)=R(1)=S(1)=0. \]](http://latex.artofproblemsolving.com/0/d/3/0d3799bfd4e92541002a6055adb144050d877405.png)
Hence, 
. 
divides 
from ,of course assuming that is not 
.
divides from ,of course assuming that is not .
![\[x^4+x^3+x^2+x+1 \mid x^5-1 \mid \left[P(x^5)+xQ(x^5)+x^2R(x^5)\right]-\left[P(1)+xQ(1)+x^2R(1)\right]\]](http://latex.artofproblemsolving.com/8/1/0/8103bd9135d49b300d5572b67ca69686c97a6da6.png)
Then![\[x^4+x^3+x^2+x+1 \mid P(1)+xQ(1)+x^2R(1)\]](http://latex.artofproblemsolving.com/4/f/d/4fdf2a5d8b5aacdb20c2f15f5f86cb639bc60622.png)
But 
then![\[P(1)+xQ(1)+x^2R(1) \equiv 0 \ \forall \ x\]](http://latex.artofproblemsolving.com/5/8/5/585729b07c902a35838dc3194a7b5b2140aa944d.png)
Hence, 
. Prove that
.
is the inverse of function 
and thus, they must intersect that 
axis (as its clearly not self-inverse function). Thus: 
is the only solution.
be a fifth root of unity not equal to 
.
and 
one at a time for 
and add up the five resulting equations.
and that 
,
while the right hand side is 
,
,
.![\[ x P(x^5) + x^2 Q(x^5) + x^3 R(x^5) = x(x^4 + x^3 + x^2 + x + 1) S(x)\]](http://latex.artofproblemsolving.com/4/b/1/4b10de4e307ba9b817eaa08286dc863ab67c0fa2.png)
for the five roots 
to get![\[ 0 = P(1)\sum_{k = 0}^{4}{\omega^k} + Q(1)\sum_{k = 0}^{4}{\omega^{2k}} + R(1)\sum_{k = 0}^{4}{\omega^{3k}} = 5S(1) = 5P(1).\]](http://latex.artofproblemsolving.com/2/6/1/26187bacf0e34bc169d1f85f2e6a800873613e0c.png)
a primitive fifth root of unity, then the five equation that result from plugging in 
for 
for 
are
.
yields 
.
then from the first two equations we have
.
,
so 
which means 
is a factor of 
as desired.
![\[\begin{vmatrix}1&\omega&\omega^2\\1&\omega^2&\omega^4\\1&\omega^3&\omega\end{vmatrix}=\omega^3+2-2\omega^2-\omega^4\neq0,\]](http://latex.artofproblemsolving.com/c/f/d/cfdae26343895428bcb86efb7bca01910870dddc.png)
one can immediately deduce that 
.
,
to the original equation, we get 
.
gives 
.
denote the given assertion, and let 
Adding up all of these yields 
.
which is more than enough. Solving the system, we easily get 
and the fact that if x-1 is a factor of P(x) suggests P(1)=0 gives suggestions to find ways with powers of 5 and 1. With experience, you will find motivation to consider the roots of unity. Now summing over 
with x=a fifth root of unity not equal to 1, and 
,
(in some order, 1, x, …, x^4 each appear once in each root of unity) 
,
,
,
