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 a function such that
for all 
. Determine all such functions 
.
lies inside of scalene triangle 
with incenter 
such that 
Lines 
and 
intersect line 
respectively at 
and 
. Line through parallel to 
intersects 
at 
and line through parallel to 
intersects 
at 
. Prove that triangles 
and are similar.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
such that for any positive integer 
there exists a positive integer 
such that the last digits of 
are equal to 
of the triangle 
which doesn't coincide with 
or 
intersectes the altitudes from 
and 
at 
and 
respectively. Let 
be the reflection of in and 
be the reflection of in 
Prove that the circles 
and 
are tangent.
of the vertices of the regular polygon 
we write the number 
and in the remaining ones we write the number 
Let 
be the number written on the vertex 
A vertex is good if ![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)
for any integers 
and such that 
Note that the indices are taken modulo 
Determine the greatest possible value of such that, regardless of numbering, there always exists a good vertex.
is the 
-excenter of the triangle and 
is the circumcircle of this triangle. 
is the middle of arc 
of which doesn't contain 
. 
meets at 
. Prove that
, point 
is marked. Point 
is on segment 
, and point 
is on segment 
such that the circumcircles of triangles 
and 
are tangent to line 
. Lines are drawn through points 
and 
passing through the center of the circumcircle of triangle 
, and through points and passing through the center of the circumcircle of triangle 
. Prove that there exists a circle tangent to all four drawn lines.
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
and distinct positive integers 
are given. Does there exist a polynomial ![$p(x)\in\Bbb{Z}[x]$](http://latex.artofproblemsolving.com/5/1/a/51a23efd70e495d2fe516b642f0a6e71e5875ba2.png)
of degree 
that satisfies the following conditions?
![$p(x)\in\Bbb{Q}[x]$](http://latex.artofproblemsolving.com/d/1/5/d152c23297a62497b4a36f3cc5955ba13e6aa920.png)
it is easy, the main problem is to be more accurate fo get integer coefficients.
it is easy, the main problem is to be more accurate fo get integer coefficients.
and 
, the Lagrange interpolation formula tells us that a polynomial ![$p\in\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/6/f/46f5a28ec6f56f7d0b8e2d49a14dceeb07a32757.png)
with degree exists, such that 
. satisfying the condition.
as above, one can use Dirichlet Theorem to find infinitely many primes 
, and let 
be the product of some primes.
, but I'll write it up.
For each 
, select a prime 
which is congruent to 
(mod ). Select these primes in such a way so that all 
of them are distinct, which is doable by Dirichlet's Theorem. By convention, we will let 
for 
denote 
![$P \in \mathbb{R} [x]$](http://latex.artofproblemsolving.com/9/7/c/97c84001b3544d2af386811b8e9ce9ccd6e641f1.png)
of degree such that:
It's easily checked that this polynomial satisfies the conditions of the problem, so all that remains is to check that it has integral coefficients. Indeed, observe that 
is an integer congruent to (mod ) for each 
Therefore, the following claim solves the problem.
of degree so that 
is an integer divisible by for each 
, ![$Q \in \mathbb{Z}[x].$](http://latex.artofproblemsolving.com/4/0/d/40d960e0e05519c246ef64ad1ed40013a6869bc8.png)
is given by:
since 
, we've shown that ![$P -1 \in \mathbb{Z}[x] \Rightarrow P \in \mathbb{Z}[x],$](http://latex.artofproblemsolving.com/8/1/5/8159c371f42bf27d3e3a55d2015057738fc594cd.png)
so we're done.
be the product of primes up to 
. We can get 
to be divisible by primes 
for 
. We want 
in 
so we need 
since 
. We select 
such that 
and 
. We can construct 
via the Chinese Remainder Theorem since 
. works. Note 
and 
. Assume for contradiction 
divides 
and 
. We know 
for all primes less than 
so 
must be injective mod . Then in ![$\mathbb{Z}_p[x]$](http://latex.artofproblemsolving.com/3/f/d/3fd6af0acc8b77146f2f801902f10d7e893644f4.png)
, 
. Since 
it follows that is the zero polynomial. However, 
so p cannot divide both simultaneously, contradiction!
. And 
. By Lagrange interpolation formula we have there exist a polynomial with rational coefficients that fits on 
with 
. We just need to ensure its coefficients are in integers; I'll choose such that 
for all 
and it's easy to see the coefficients of will be integers. For this just induct on . we will prove for all positive integer ; there exist an arbitrary large sequence(not infinite) like such that that 
and 
and 
. For adding new element to the sequence like 
just choose arbitrary large( larger than any previous elements) distinct primes like 
such that 
. Just set 
for 
and 
. You can see this sequence of 
satisfies the conditions...
and 
.
such that:
such that 
.![$P(x) \in \mathbb Q[X]$](http://latex.artofproblemsolving.com/0/a/e/0ae67d89d989c0f6a1be711bda3d9db41191ec72.png)
with degree atmost ![$P(x) \in \mathbb Z[X]$](http://latex.artofproblemsolving.com/0/a/4/0a4bf0bc0c69b822b6b555351731e9cc40169945.png)
relatively coprime integer congruent to 
works well .
![$P(x) \in \mathbb Z[X] $](http://latex.artofproblemsolving.com/3/a/9/3a9dd2fdf39a54b296fe2fb9943fc3de5a3810d4.png)
.
