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.![\[ f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n \]](http://latex.artofproblemsolving.com/0/9/d/09d80fe5c25c5c6ea57c545b68c7e11b319cbada.png)
with integer coefficients 
, and given also that there exist four distinct integers 
, 
, 
and 
such that ![\[ f(a)=f(b)=f(c)=f(d)=5, \]](http://latex.artofproblemsolving.com/8/2/d/82d0c5d0c7a6dd48802ad8fbe04b211278c1fa58.png)
show that there is no integer 
such that 
.
be positive real numbers so that 
. Prove that![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
has different prime divisors. Prove that there exists a positive integer number 
such that 
divides by .
st sees the other three, the 
nd sees the 
rd and 
th, and the rd sees only the th. The devil has 
hats, numbered from to , he puts one hat on each wise man, and hides the extra 
hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
be an acute triangle with circumcircle 
. Let 
be the midpoint of 
and let 
be the midpoint of 
. Let 
be the foot of the altitude from 
and let 
be the centroid of the triangle . Let 
be a circle through and that is tangent to the circle at a point 
. Prove that the points 
and 
are collinear.
be the incentre of acute-angled triangle . Let the incircle meet 
, and at 
, and 
respectively. Let line 
intersect the circumcircle of the triangle at 
and 
, such that 
lies between 
and . Prove that 
. be an acute-angled triangle with circumcircle . A circle 
is internally tangent to at and also tangent to 
at . Let and intersect at and respectively. Let 
and 
be points on line such that 
is the midpoint of 
and 
is the midpoint of 
. Lines 
and 
meet at 
and intersect again at and 
respectively. The ray 
meets the circumcircle of triangle 
again at 
.
. the circumcircle of the acute triangle . Point is the midpoint of the arc of not containing . Circle centered at is tangent to the segment at point . Tangents to the circle passing through point intersect line at points and 
such that points 
lie on the line in that order. Circle 
is tangent to the segments 
and 
and to the circle at point . Circle 
is tangent to the segments 
and 
and to the circle at point . Lines 
and 
intersect at point . Prove that 
.
be the set of all positive integers. Find all functions 
such that 
divides 
for all positive integers 
and .
be a positive integer. Dominoes are placed on a 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each , determine the largest number of dominoes that can be placed in this way.
or 
. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
and 
are given. Losyash likes some of the cells of the 
checkerboard. In addition, he is interested in any checkered rectangle with a perimeter of 
, the upper-left corner of which coincides with the upper-left corner of the board (there are such rectangles in total). Given and , determine whether Losyash can color each cell he likes in one of colors so that in any rectangle of interest to him the number of cells of any two colors differ by no more than .
is the midpoint of side of an acute—angled triangle . The point 
is symmetric to the orthocenter relative to its circumcenter. The point 
inside triangle is such that 
. Prove that 
.
is given. Suppose that 
, 
and 
are different positive integer numbers such that 
. Prove that 
.
and 
denote two sequences of integers defined as follows:
Prove that, except for the "1", there is no term which occurs in both sequences.
and the same for 
then,
![\[x_n=X(-1)^n+Y2^n.\]](http://latex.artofproblemsolving.com/b/6/4/b64103ce6b36840afb72e204ce9d8bde7c3ae59e.png)
Using 
and 
yields ![\[x_n=\frac{2^n+(-1)^{n+1}}{3}.\]](http://latex.artofproblemsolving.com/0/e/9/0e91ecb368a2adce3368cc089356ac41c89c7f0a.png)
Similarly, solving for the second recurrence yields ![\[y_n=2\cdot 3^{n-1}+(-1)^n.\]](http://latex.artofproblemsolving.com/b/e/9/be9151fca1e0c704e75c410c3d6963099b9170e1.png)
So if 
then 
or 
.
or , then 
is the only solution, corresponding the the fact that the term is in both sequences. If 
, then 
. But we have 
, so ![\[2\cdot 3^n+3(-1)^n+(-1)^m\equiv 5(-1)^n+(-1)^m\pmod 8\]](http://latex.artofproblemsolving.com/c/9/6/c965e92b534b4e7746ecd90274be64ceaa3d0122.png)
which cannot be 
. Hence there are no solutions for , and the only integer in both sequences is indeed . 
cannot be equal to 
. 
.
. Now suppose this is true for some . Since 
, and 
, so adding these two gives 
, which means will always be greater than for all . 
and the same for 
then we would have 
.
,
and 
,
,
.
.
,
,
or 
,
.
.
.
if 
,
.
where 
and 
are roots of the equation 
.
and 
we get 
.
.
we get 
.
if 
, so we get 
.
we get 
which is not true hence , sequence 
and 
shares no common numbers other than 