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.
such that
for all 
.
be a triangle with 
. Let 
be the midpoint of 
, and let 
be a point on segment such that 
. Let 
be the point of intersection, different from 
, of the circumcircle of triangle 
and line 
. Define 
and 
as the points of intersection of line 
with 
and 
, respectively. Prove that is the midpoint of 
.
satisfies 
. If 
, find the maximum value of 
.
with length-side 
, prove that
satisfies 
. Point 
lies inside so that ![\[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\]](http://latex.artofproblemsolving.com/6/d/7/6d798ef2bf8136054abb232961c3b646e7f320a4.png)
Prove that 
.
+1 w such that for all real 
![\[f(x+f(y))=f(x+y)+y\]](http://latex.artofproblemsolving.com/1/e/2/1e2fef35ee53cd3224527f672b9edbb7a5293fad.png)
and 
such that 
divides 
and 
divides .
and two points 
and outside , a quadrilateral 
is defined as "good" if 
are four distinct points on in order, and lines 
and 
intersect at and lines 
and 
intersect at .
, let 
denote its area. If there exists a good quadrilateral, prove that there exists good quadrilateral such that for any good quadrilateral 
, 
.
be an odd integer, 
. Consider 
integers 
such that for any 
, 
and 
. Prove that the number of 
for which satisfy 
for some 
, 
is greater than 
.
be a positive integer and 
be a polynomial with integer coefficients such that 
.
is divisible 
.
and 
be two infinite sequences of integers. Prove that there exists an infinite sequence of integers 
such that for any positive integer , the following holds:![\[
\sum_{k|n} k \cdot z_k^{\frac{n}{k}} = \left( \sum_{k|n} k \cdot x_k^{\frac{n}{k}} \right) \cdot \left( \sum_{k|n} k \cdot y_k^{\frac{n}{k}} \right).
\]](http://latex.artofproblemsolving.com/2/5/9/259ee32c89befe7a3ee2797e0d5b1df1c3066cf6.png)
has degree 
. Clearly, we may assume that the leading coefficient of is positive. In that case 
for all where 
is a suitable constant depending only on and hence![\[f(n+1)+\dots+f(n+k) +Ck \gg (n+1)^d+\dots+(n+k)^d \gg (n+1)^2+\dots+(n+k)^2 \gg kn^2+k^3.\]](http://latex.artofproblemsolving.com/9/1/e/91ead1ddad5156a80dc12120515b9a98888c4b78.png)
So if we want to represent an integer 
, we need 
and hence 
and 
.
is only 
so that almost all the integers up to 
are not represented, if is very large, in particular, there are infinitely many such numbers. is 
by counting slightly more carefully, but this is irrelevant here.)
is positive. Let 
be sufficiently large positive integer. The plan is to count the possible integers in the interval ![$[N,2N]$](http://latex.artofproblemsolving.com/4/6/9/4698e25eee241a60a09c070774a401128920fc03.png)
that can be represented as required. Note that
for sufficiently large integers , where 
is a positive constant, that may depend on . By 
it follows
where 
is a positive constant (possibly depending on )
. The number of integers in 
that can be represented as sum of at most 
consecutive values of is at most 
. Let now estimate the number 
of integers ![$r\in[N,2N] $](http://latex.artofproblemsolving.com/6/9/b/69bebb5ac6a8fbc1feee78d02509ccd1c60ab1be.png)
representable as
It means 
, thus 
or
which holds if is sufficiently large . Further
Hence the number of integers in that can be represented as required is at most
which is less than for sufficiently large (and depending 
).
by 
, everywhere, having in mind the Prime Number Theorem, seems enough?) Is there are way to prove this in a purely number-theoretic fashion? (I have a good approach but with an awful gap, may post later.)
. Note 
is a polynomial of degree 
. Let 
, 
, then 
then
is bounded. If 
this limit is 0.
for a suitable 
. This means 
so this produces 
solutions.
, then 
so 
. For each choice of 
, we are dealing with a polynomial 
, and each of them covers 
numbers in ![$[1,n]$](http://latex.artofproblemsolving.com/2/2/4/2240bb122bdadbe086c0a6291dcd5a3eb82ad5cf.png)
so they cover at most 
numbers.
. We may assume that 
has positive leading coefficient (otherwise it is bounded from above for 
), thus 
as 
and so the sets 
, 
cover the positive integers. Now, for a fixed , the number of solutions to 
(and hence the number of choices for and 
) is at most 
for some constant 
; but 
for and so we are done.
, what are they exactly? Can someone elaborate please? I couldn't find a solution sadly
for some positive integers and .
be the set of positive integers which cannot be represented in this form. Since is finite, there exists a positive integer such that 
.
. Since has degree greater than 1, 
has degree at least 1. Thus, by the integer root theorem, there exists an integer such that 
.![\begin{align*}
\sum_{i=n+1}^{n+k}g(i)&=\sum_{i=n+1}^{n+k}[f(i+1)-f(i)] \
&=f(n+k+1)-f(n+1).
\end{align*}](http://latex.artofproblemsolving.com/5/0/c/50c5e08d0ce3563f732e79028ea29cb1cac36f67.png)
is precisely the set of possible values of 
., there exist positive integers and such that 
. Then, the set of possible values of contains all the positive integers greater than or equal to 
.
, there exist positive integers and such that 
. Thus, 
, which contradicts the assumption that . Therefore, there must be infinitely many positive integers which cannot be represented in the form for any positive integers and .
is positive. Let 
, so we want to show that there are infinitely many positive integers not representable as 
for some .![$[1,N]$](http://latex.artofproblemsolving.com/5/0/c/50cbe56b7b74ba826b20c3f1217e6c3425530d4c.png)
and suppose ![$f_k(n) \in [1,N]$](http://latex.artofproblemsolving.com/c/0/f/c0f97ac4da2b2a6188d2394c2e6ec16b58124390.png)
. This clearly implies that is 
. Furthermore, we can check (by estimating with an integral, for instance) that if 
, then 
, hence must be 
, so there are 
pairs that could possibly work. 
has degree 
. We may assume that the leading coefficient 
of is positive, otherwise is bounded from above for and we are done. Let be a constant such that 
for all 
. There are finitely many integers in the desired form with 
and 
, so it suffices to show that there are finitely many with 
or 
. Consider all integers in the interval ![$[1,M]$](http://latex.artofproblemsolving.com/b/a/2/ba2114a65b72af32f9da4dfb8fc84de0e90a8903.png)
. In order for an integer in this interval to be representable in the desired form with corresponding and , we must have 
. On the other hand![\[f(n+1)+\cdots+f(n+k) \geq \frac{a}{2}\left((n+1)^d+\cdots+(n+k)^d\right) \geq \frac{a}{2}\left((n+1)^2+\dots+(n+k)^2\right)\]](http://latex.artofproblemsolving.com/5/c/6/5c667ac97ff0d49500f8701a8cca22f4c26b913c.png)
and hence 
and 
(as 
).
, we need 
and 
for some constant 
. Hence, the number of choices of is at most 
and so at least 
integers are not representable in the desired form. As becomes very large, the latter expression grows arbitrarily large, so we are done.
has degree 
. Now I will count the number of values 
that we can obtain. We can clearly assume the leading coefficient of is positive. Let the leading term of be 
. Clearly we have that 
. If 
as we have the number of values we can obtain from the first values is 
we get that for large enough we can make 
arbitrarily large which suffices. Now suppose that 
. Let 
be the number of contiguous substrings 
, 
, 
, 
of 
, 
, , such that 
. As , we get that for sufficiently large values of 
as when is large we get that relation when 
and as this means we get as gets large. Also note that 
. If we let 
be the number of values that we can obtain that are . Clearly we have that 
. Thus as and for all 
. If we take a very large value of we can make 
arbitrarily large as 
.
and 
Prove that 
![\[f(n+1)+f(n+2)+\cdots+f(n+k)\]](http://latex.artofproblemsolving.com/a/6/4/a648623198555fa749157c99a83eec8e74c98961.png)
where 
"
