[/quote]
,
be reals such that 
and 
Find the value of 
be a triangle with circumcentre 
.
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
.
lies on the line 
and 
and 
on that line.)
be the circumcentre of triangle 
.
and 
,
to tangent 
disjoint from 
on 
and 
.
is tangent to two fixed circles.
as the set of all products of an element of 
and 
then 
We call a set 
of positive integers good if there do not exist sets 
of positive integers, each with at least two elements and such that the sets 
and 
is good.
,
and 
are integers.)
be the midpoint of side 
and 
be the feet of the altitudes from 
and 
intersect at a point 
is perpendicular to line 
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
.
and 
with conditions,
is a prime number.
is a perfect square.
is also perfect square.
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
such that:
for all 
for all 
is a natural number. A function 
is called \textit
-
with 
,
has a solution in natural numbers 
such that ![\( \frac{y_0}{x_0} \in \left[\frac{1}{100}, 100\right] \)](http://latex.artofproblemsolving.com/d/a/6/da67f464a934f890626d188b84940fc82c4f4ff9.png)
,
is a solution. Suppose 
,
is a polynomial with real coefficients, negative leading coefficients, and total degree greater than 2, and for every real number 
,
as ![\( \frac{y}{x} \in \left[\frac{1}{100}, 100\right] \)](http://latex.artofproblemsolving.com/6/5/d/65d4154dd5fb612f92f24f98406bc9bcfc713806.png)
.
is not 
such that for all 
be defined by![\[
x_1 = 2,\quad x_{n+1} = x_n + \frac{n}{x_n},\quad \text{for all } n \geq 1.
\]](http://latex.artofproblemsolving.com/f/2/8/f28193d8ea3fa6da8bbdb51f9a7e2e9d8a7d3c9f.png)
Prove that the sequences 
and 
have finite limits, and find those limits.
Y by Davi-8191, Mathuzb, Epistle, megarnie, microsoft_office_word, Adventure10, Mango247, Sedro, Gato_combinatorio, and 1 other user
Call admissible a set 
of integers that has the following property:
If
(possibly 
) then 
for every integer 
.
Determine all pairs
of nonzero integers such that the only admissible set containing both 
and 
is the set of all integers.
Proposed by Warut Suksompong, Thailand
of integers that has the following property:If
(possibly 
) then 
for every integer 
.Determine all pairs
of nonzero integers such that the only admissible set containing both 
and 
is the set of all integers.Proposed by Warut Suksompong, Thailand
This post has been edited 3 times. Last edited by WakeUp, Jan 4, 2014, 3:51 AM
Reason: Edited Title and Changed Format of The Problem.
Reason: Edited Title and Changed Format of The Problem.
then the set of all multiples of 
satisfy the question too, hence this pair isn't one we want.
then by 
we can get all multiples of 
and similar for 
there exist 
such that 
with 
and 
.
,
)
are involved, one can smell Bezout. However, the condition is a quadratic one, not a linear one, so the trick is to surpass this obstacle, by introducing some expressions of degree 
that belong to 
an admissible set. We will try to prove 
.
.
,
.
,
.
such that 
.
and 
for all 
integers. By Bezout, we can set 
then every element of the set 
satisfayes the conditions.Now,if 
that every integer of the form 
and 
is in the set,so we can pick 
(
and 
and 
.
.
will work.
,
will be in 
.
we will have every multiple of 
.
.
.
and 
in 
we have 
So, we can always form 
and taking 
we will have every integer in 
then 
is possible.
,
.
.
as desired.
then 
so the set 
is clearly admissible because every element must be divisible by 
then by choosing 
,
so every multiple of 
is in 
so every multiple of 
is in 
.
so that 
.
.
.
is the set of all integers if an only if 
such that 
then all the elements of the admissible set containing 
so admissible set containing 
for all 
.
for all 
Also note that taking 
we obtain that 
Therefore, it means that 
for all 
Now, as 
so by Bezout's identity there exists 
such that 
thus we obtain that 
Choosing 
it follows that 
And we are done. 
which satisfy the problem condition are precisely those for which 
.
are pairwise relatively prime.
,
.
.
,
.
,
.
,
are members of 
are members of 
,
such that
For such an 
are members of 
.
for all 
,
,
,
.
to get that 
,
,
that 
.
,
,
.
and 
,
,
.
,
.
, then we have that we only generate pairs of numbers such that they are divisible by 
divides every element of 
denote the assertion 
.
,
for any 
,
such that 
gives![\[ 1=(am^2-bn^2)^2 \in A. \]](http://latex.artofproblemsolving.com/5/0/3/5032fb5f1c253ae3cb76e90255237560291caa6e.png)
Finally, 
gives 
for any 
.
,![\[g^2 \mid x^2+kxy+y^2\]](http://latex.artofproblemsolving.com/b/d/b/bdb1901dcb96999d88b4b1129b67cbb9e2b7b929.png)
Thus, we cannot generate the necessary infinitely many elements not divisible by 
.
,
.
,
.
with ![\[(xm^2-yn^2)^2 = 1\]](http://latex.artofproblemsolving.com/0/3/6/0364d791c60f461ea93080b72bd4fa5bf261acda.png)
.![\[ax^2-by^2=1.\]](http://latex.artofproblemsolving.com/5/c/1/5c1c48ce549ee6f379c81564a92425ba4e15c78d.png)
Now let 
and 
.![\[m=x^2+k_1x^2+x^2=((2y^2-2x^2+1)a+2)x^2\in A\]](http://latex.artofproblemsolving.com/4/8/5/485d2aebf7d766196fd38abba8e7c02203839d76.png)
and![\[n=y^2+k_2y^2+y^2=((2y^2-2x^2+1)b+2)y^2\in A.\]](http://latex.artofproblemsolving.com/5/f/2/5f2af2ca8263fbf946876967ca3408329f63f3d9.png)
But 
,![\[m^2+(-2)(m)(n)+n^2=(m-n)^2=1\in A.\]](http://latex.artofproblemsolving.com/5/7/d/57d9f67d0e5f46666696604f1f026d0d74db81a5.png)
Now for any positive integer 
![\[(1)^2+(z-2)(1)(1)+(1)^2=z\in A\,\blacksquare\]](http://latex.artofproblemsolving.com/0/e/f/0ef04b20718bf5100f121f0e3111fc35561b07c9.png)
are coprime, then all multiples of 
lie in the set from 
and 
.
such that 
.
we must have 
,
we have all integers must lie in 
.
.
finishes.
such that 
work. Clearly 
and 
show that all multiples of ![\[am^2 - bn^2 = 1;\]](http://latex.artofproblemsolving.com/b/4/b/b4b4c3b7a0bb68344bbcbda8975dca679af7c753.png)
taking 
and 
or 
results in 
covers all the integers. Now, let 
so that 
and 
.
,
,
.
and 
,
all multiples of 
and 
.![\[r^2-2r(r+1)+(r+1)^2=r^2-2r^2-2r+r^2+2r+1=1\in A\]](http://latex.artofproblemsolving.com/f/1/b/f1b1bef960a2371c7eb0cef6961dbefab131521b.png)
and thus, by our first observation, 
,
the multiples of 
and 
all multiples of 
gives 
Hence, 
.
,
then we can see that the set 
works.![\[x^2+kxy+y^2 (1) \]](http://latex.artofproblemsolving.com/0/2/3/02366e6b0829b957faab0b3911fa8011cba6b4c0.png)
We will have all multiples of 
then 
,
as desired.
which is undesirable(since we won't be able to get numbers that are not divisible by 
.
we can plug in 
with 
to get 
with 
.
for all integers 
and 
.
.
so 
.
are in the set. We can then find two of these numbers with difference 
and we get that 
are part of the set, which is just all integers.
then obviously every element of 
by using 
in the given condition, we see that every multiple of 
and similarly every multiple of 
Since 
so by Bezout we may find integers 
such that 
Since 
we see that 
Then taking 
in the given condition, we see that every integer is in 
Thus letting 
yields 
Now letting 
for all integers 
,
for every integer 
for any integer 
,
for every integer 
)
)
.
and 
are in an admissible set then so is 
,
is also in 
which implies that 
.
also belongs to 
since 
by putting 
,
proving the lemma.
.
works well , while we note that 
are there; select 
to get 
, set 
,
.
to get that 
is the only possible admissible set(obviously 
