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.![$f \colon [0,1] \to \mathbb{R} $](http://latex.artofproblemsolving.com/2/1/8/21859d37ed035abd07d0be112f49d52598fa477e.png)
be a differentiable function such that its derivative is an integrable function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
, and 
. Prove that ![\[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]](http://latex.artofproblemsolving.com/2/3/e/23e33eed39fc1e20486bbb86261d3566e8fd2017.png)
be a positive integer. Prove that there exists a positive integer 
such that exactly 
of the integers 
are squarefree.
which verify relation![\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/3/a/2/3a235e8be4c61c32f376999e61d64973db21dc75.png)
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
for which there is a positive real number such that![\[ f'\left(\frac ax\right)=\frac x{f(x)} \]](http://latex.artofproblemsolving.com/b/1/4/b1428d1e639f7c515404d6f9e37e027a6ec82abe.png)
cells of an 
square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell 
, one can reach any other black cell 
through a sequence 
of black cells where each consecutive pair 
are adjacent (sharing a common side) for every 
. Let 
denote the maximum possible cost over all initial colorings with exactly black cells. Determine a constant 
such that![\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]](http://latex.artofproblemsolving.com/0/a/0/0a06d4ff52479c8cb785c6450cea8748bf7769b6.png)
for any 
.
+1 w
be the orthocenter of triangle 
. Let 
be the circumradius of . Prove that triangle is equilateral iff
be a circle, 
be a diameter of the circle and for each point 
let 
be the foot of the perpendicular from on 
and 
the base of the perpendicular from 
on 
. Let 
be the geometric locus of all points if passes through all points on the circle . Prove that actually has no axes of symmetry other than , as shown in the picture.
and be positive integers. In an 
grid, two cells are considered neighboring if they share a common edge. Kritesh performs the following actions:
in any cell of the grid.
. and .
, s.t :
of real numbers such that![\[ a_{m^2 + m + n} = a_{m}^2 + a_m + a_n\]](http://latex.artofproblemsolving.com/b/b/9/bb97053e86e7733fe5c26f7859c87566eb444b52.png)
for all positive integers and .
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
, and let 
for 
. Determine whether![\[ \sum_{n=1}^{\infty}a_n^2 \]](http://latex.artofproblemsolving.com/5/4/0/540a4f3740d9b561c40dcc36ea9350d94421dd86.png)
converges.
as an example. Each eigenvalue 
of is a root of 
which is irreducible by Eisenstein. And also because it has no rational root. Also ![$[Q(t):Q]\le 2$](http://latex.artofproblemsolving.com/e/0/a/e0a35fdece3a667234bf58b762cd33c123b6b812.png)
since 
has degree 2. Contradiction. Hence for we have no solution in 
. The equation is solvable iff is the cube of an integer.
where 
Hence 
, and consequently also , is the cube of an integer. not divisible by 
., however, the equation is always solvable.
doesn't divide , then 
and 
.
If 
then 
and 
, where 
.
so that is an integer matrix ? (In general, the integer matrices are not sufficient). If 
then is diagonalizable over 
and 
, where 
.
. so that is an integer matrix ? divides and 
, then 
, is diagonalizable over and 
. is not a cube, then 
is irreducible over 
and we can group the eigenvalues by , that is, 
.
, where 
. so that is an integer matrix ?
is solvable over 
for every 
.
,
.