[/quote]
,
be an odd integer. Determine all functions 
from the set of integers to itself, such that for all integers 
and 
the difference 
divides 
for which there exist infinitely many positive integers 
such that ![\[ \frac{a^m+a-1}{a^n+a^2-1} \]](http://latex.artofproblemsolving.com/2/8/a/28a68e409451bbdcaee7c1286341f6aeea9ffdad.png)
is itself an integer.
of positive integers such that 
is multiple of 
and 
is multiple of 
.
be positive integers such that 
is a multiple of 
for all positive integers 
.
of positive integers such that 
is divisible by 
.
be a circumscribed quadrilateral and let 
be the intersection point of its diagonals 
and 
.
be the incenters of 
,
,
,
,
intersects the sides 
and 
at points 
,
intersects the sides 
and 
at points 
and 
,
such that there exists a pair 
is not divisible by the cube of any prime, and 
,
be a variable point on the line 
such that 
lies between 
and 
and 
intersect at two distinct points 
and 
Prove that the line 
passes through a point independent of 
be a triangle with circumcircle 
and incentre 
.
intersects the lines 
,
,
at points 
,
,
,
,
,
,
of the segments 
,
,
.
and 
. Prove that
Where 
,
,
and the line 
and the line 
and 
meet at 
.
of the triangle 
.
and circumcenter 
.
.
be the orthocenter and 
of the circumcircle of triangle 
,
.
be the circumcenter of triangle 
.
and 
meet on 
,
.
with diameter 
,
at a point 
.
at 
and 
intersect at a point 
on 
.
Y by anantmudgal09, rkm0959, adityaguharoy, MonsterS, Davi-8191, tenplusten, Rounak_iitr, Adventure10, Mango247, Deadline
Determine all positive integers 
satisfying the following condition: for every monic polynomial 
of degree at most with integer coefficients, there exists a positive integer 
and 
distinct integers 
such that ![\[P(x_1)+P(x_2)+\cdots +P(x_k)=P(x_{k+1})\]](//latex.artofproblemsolving.com/e/e/8/ee8adc79dc258b021529245fec16e081da2d1b0a.png)
.
Note. A polynomial is monic if the coefficient of the highest power is one.
satisfying the following condition: for every monic polynomial 
of degree at most with integer coefficients, there exists a positive integer 
and 
distinct integers 
such that ![\[P(x_1)+P(x_2)+\cdots +P(x_k)=P(x_{k+1})\]](http://latex.artofproblemsolving.com/e/e/8/ee8adc79dc258b021529245fec16e081da2d1b0a.png)
.Note. A polynomial is monic if the coefficient of the highest power is one.
This post has been edited 1 time. Last edited by IstekOlympiadTeam, Feb 25, 2017, 5:45 PM
fails. However, to show that 
works, note that linear polynomials work; for quadratic polynomials, if 
is the middle term, then 
,
fail.
fail, let 
.
if 
is odd, and 
if 
have odd degree). It is easy to check that this has degree 
for 
.
,
.
,
.
cannot happen, but the derivative is positive everywhere, so we are done.
fails, just take 
.
,
for all 
fails, take 
.
,
.
,
.
,
.
is an integer, 
,
only.
(take 
(take 
)
,
are never injective, and hence fail the constraint when 
.
.
does not work. Select a prime 
such that 
(by Bertrand postulate). Then, consider the polynomial![\[ P(x) = x^p + (2p-1)x + 1. \]](http://latex.artofproblemsolving.com/c/6/e/c6ebaf947f4d60caf768487807620953ec68b98b.png)
We have 
and 
for every 
,
.
exist we have![\[ \underbrace{1 + \dots + 1}_k \equiv 1 \pmod{2p} \]](http://latex.artofproblemsolving.com/9/9/c/99c84163bf7298cd9eccfd52325aa5b720d6ea5b.png)
and so 
,
we conclude 
such that 
,![\[ \frac{a^p - b^p}{a-b} = -(2p-1) \]](http://latex.artofproblemsolving.com/6/4/b/64bd9907f9bd5d4182c2c8fad02f9843e5755042.png)
which is impossible since the left-hand side is always positive (for 
the minimal 
for which there exist distinct integers 
satisfying 
.
,
.
,
follows.
the polynomial![\[
P_m(x)=x\cdot (x^2+1)(x^2+2)\dots(x^2+m)+1
\]](http://latex.artofproblemsolving.com/6/3/c/63c13b75762f6ac3001abf3391c18487eb0738cd.png)
is increasing and always 
mod 
,
.![$\cup_{m\ge 3}[2m+1;m!]$](http://latex.artofproblemsolving.com/a/c/0/ac0aa138bebb1f6ddc328260a1d6ee01b924bc7e.png)
covers 
,
does not satisfying the condition.
which is increasing and always 
,
,
,
injective and 
always, we are done. The 
.
,
.
,
.
.
,
.
always, since 
.
condition is gone. Clearly, 
,
now. Anyways, brute force calculations (very boring tbh) gives us that 
,
is a prime. Thankfully I checked this with a computer afterwards, it works 
,
.
.
.
,
.
,
as the first element of 
,
is even. This can be done because 
is odd. Anyways, after selecting 
,
where 
is injective on the integers and hence is a counterexample for 
and 
.
but only for odd 
.
.
is injective, I solved it for even 
such that![\[
P(x_1)+P(x_2)+\dots+P(x_k)=P(x_{k+1}).
\]](http://latex.artofproblemsolving.com/c/d/d/cddefa5bfe1ce01ee4be23cd2dbea44ceaa0292d.png)
Then 
.
of 
is a multiple of 
,
with 
.
be an integer with 
.
is a multiple of 
.
.
is symmetric with 
, it doesn't work.
,
( it exists because n is odd number)
can't satisfy
odd number fails
,
works the same
only satisfy
is bad. The idea is to control ![\[P(x) = (x^p - x + 1) + pQ(x)\]](http://latex.artofproblemsolving.com/6/4/4/644a2cd4aafde6d8e7789678bf4963a5a5c5f12c.png)
where 
is an integer polynomial and 
for all 
,
.
,
.
too and thus 
hence 
for which 
is clearly injective - impossible. Hence, for 
t
there exists an odd 
such that 
works
is divisible by 
let it 
we can check it manually
choose such 
is the lemma and consider
)
(consider it 
)
let 
(consider it 
)
,
.
be a sufficently large multiple of 
.
,
by, among other things, Bertrand's. Or that if 
,
is less than 
is an even function and 
.
.
so 
for all 
for all 
,
.
.
,
for all real 
works for 
works. Again 
,
(not by derivatives but by comparing 
and 
for 
)
and 
and 
(both for 
forces 
works for 
and everything else by simple bounding, implying that we can kinda throw terms together and bound as well.
.
holds. If 
.
,
,
,
,
such that the polynomial 
is injective over the real numbers for any real number 
.
has no real roots, which would imply there don't exist reals 
with 
.
is an even degree polynomial with positive leading coefficient, so it has a minimum 
.
gives that![\[Q_C'(x) > c + C > c + r > c + |c| > 0\]](http://latex.artofproblemsolving.com/d/d/e/dded858cd0aa14b374cf7651c82f2fce5a268288.png)
for real numbers ![\[ P(x) = (x-1)(x-2) \cdots (x-m) + N \cdot m! \cdot x + 1\]](http://latex.artofproblemsolving.com/9/9/b/99b6f57a33f76f2d439148e38d5f36726c091869.png)
and 
.
,
for any integer 
and 
with 
,
,
,
.
,
or 
fail for any odd integer 
,
,
,
for any 
Since 
for all 
which is impossible unless 
cannot hold. Hence we wish for
to never hold, which is true since the LHS is never negative (this is where the motivation to put in the additional 
comes from).
holds solutions for all degree 
,
.
,
for any integer 
,
be the prime factorization of ![\[R(x) = \left(x^{p_1^{k_1}-p_1^{k_1-1}+k_1} + \left(p_1^{k_1} - 1\right)x^{k_1}\right)\left(x^{p_2^{k_2}-p_1^{k_2-1}+k_2} + \left(p_2^{k_2}-1\right)x^{k_2}\right) \cdots \left(x^{p_\ell^{k^\ell}-p_\ell^{k_\ell-1}+k_\ell} + \left(p_\ell^{k_\ell}-1\right) x^{k_\ell}\right).\]](http://latex.artofproblemsolving.com/f/a/9/fa99db05cd2924290c1bb8ed1f928c6114e73ec0.png)
Otherwise, let 
,
to 
.
is currently even, we append a factor of 
to 
.
.
for all positive integers 
,
and 
when 
.![\[\deg P - 1 \leq p_1^{k_1} + p_2^{k_2} + \cdots + p_\ell^{k_\ell} \leq p_1^{k_1} p_2^{k_2} \cdots p_\ell^{k_\ell} = n\]](http://latex.artofproblemsolving.com/e/4/9/e495613027932f7a62cc5cfd90b0cb47f0347dbf.png)
because each power is at least 
is a prime or 
.
,
is a multiple of 
.
,
,![\[\nu_p\left(m^{p_i^{k_i} - p_i^{k_i-1}} - 1\right) = \nu_p\left(m^{p-1} - 1\right) + \nu_p\left(p^{k_i-1}\right) \geq k_i\]](http://latex.artofproblemsolving.com/0/2/f/02f37c85438dbf913728477f8d5d22065c756b2e.png)
by LTE; thus 
for any positive integer 
,
mod 
monotonically decreasing for 
and monotonically increasing for 
,
is even for all 
)
,
for all positive integers ![\[(a-b)\left((a+b)\left(a^2+b^2+15\right) - 24\right) = 0.\]](http://latex.artofproblemsolving.com/3/9/4/394109d9f7bb0551ab121df2bbcb4b171f10befb.png)
We can check that there are no pairs of integers 
such that 
by a finite case check. This finishes the proof.
,
.
.![\[k \equiv P(x_1) + \dots + P(x_k) = P(x_{k + 1}) \equiv 1 \pmod{n}.\]](http://latex.artofproblemsolving.com/a/e/7/ae78cfcddde6d9a9ccdcaf4e6123a558e430eb7a.png)
Since 
this implies 
.![\[\frac{P(x) - P(y)}{x - y} = \frac{x^{\varphi(n) + 1} - y^{\varphi(n) + 1}}{x - y} + n - 1 \geq n - 1 > 0\]](http://latex.artofproblemsolving.com/8/0/c/80c5f9b7bcaecf1592bdb6b610b11b0586a67c29.png)
which implies 
,
we can simply pick distinct integers 
such that 
,
.![\[P(z) = (x-y)(x + y + a) = x^2 - y^2 + ax - ay = P(x) - P(y) \implies P(x) = P(y) + P(z)\]](http://latex.artofproblemsolving.com/a/9/f/a9fb39df1db2e85e439036dfa5ed34d46a42dae2.png)
as desired.
,
and choosing 
makes both 
and 
integers. If 
,
,
,
,
,
when 
.
,
.
,
)
when 
and 
if 
is only achieved with 
)
if 
(and 
,
and for 
,
,
and 
,
happens iff 
,
yields 
)
