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.
and 
are prime numbers. Solve each equation separately:
. Let the points 
and 
be on the sides 
, respectively, so that 
, and 
passes through the incenter 
. Let 
meet 
at 
, 
meets 
at 
and 
meets 
at 
. Prove that the line 
passes through the midpoint of 
.
be positive real numbers. Prove that
of integers is defined by![\[
-\frac{1}{2} \leq a_{n+1} - \frac{a_n^2}{a_{n-1}} \leq \frac{1}{2}
\]](http://latex.artofproblemsolving.com/3/3/b/33b59af16d750e3e7ab625a435013fa3a856e67a.png)
with 
, 
, prove that 
is odd for all values of 
.
and 
be integers such that 
and 
. Place rectangular tiles, each of size 
, or 
on a 
chessboard so that each tile covers exactly cells and no two tiles overlap. Do this until no further tile can be placed in this way. For each such and , determine the minimum number of tiles that such an arrangement may contain.
is the expression 
an integer?
such that![\[f'(x)=\frac{f(x+n)-f(x)}n\]](http://latex.artofproblemsolving.com/8/5/7/857920b9c714ca96bb50dfdebcad18d37287db99.png)
and all positive integers 
of real numbers.
grid 
. In the first row of this grid, we place 
in every square (
. In the first column of this grid, we place in the -th square (
.
.
for all 
, does this mean that 
for all ?
be any positive integer not equal to 
or 
. Show that one can find distinct 
in the set 
such that 
is not a perfect square.
for 
.
, then according to the Leibniz Integral Rule we have
and 
functions are given with following properties:
is strict increasing and for each 
there holds 
or 
.
if holds and 
if holds.
and 
such that for all 
we have 
does there exist an matrix 
whose entries are all in 
, such that 
is the matrix of all ones? has 
times the eigenvalue 0 and the eigenvalue . So the charaterisric polynomial of must be 
. This is only for 
an integer polynomial.
and let 
be the 
-matrix with entries 
. We can choose as the block-matrix 
.
.![$Z=[0,\cdots,0]^T,V=[1,\cdots,1]^T\in \{0,1\}^m$](http://latex.artofproblemsolving.com/0/5/b/05be892fde5a92fb2627f6f1384b3e5e18ae6edf.png)
; then put![$M_0=[V,Z,\cdots,Z],M_1=[Z,V,Z,\cdots,Z],....,M_{m-1}=[Z,\cdots,Z,V]$](http://latex.artofproblemsolving.com/4/5/b/45b6c7cb39f4ac60ce4bbec97cbbb9f41bf44509.png)
.
.
and 
.
is a perfect square.
and pairs 
(e.g. via digits in 
iff for the corresponding pairs 
and 
we have 
.
for all 
.
such that 
.
.
.
.
,
,
for some integer 
hence 
is a perfect square.