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.
, in which case the object is not distributed).
derives a payoff of 
when he wins object 
. If he wins half of , then his payoff is 
. The values 
are given beforehand.
as the indicator function that is assigned the value 
if player entirely wins object , value 
if player wins half of and otherwise.'s total payoff after the game concludes is given by 
where 
is the set of five objects. That is, the sum of payoffs over all objects won.
. Player 's strategy bid-vector is given by ![$B_i = [b_1, b_2, b_3, b_4, b_5]$](http://latex.artofproblemsolving.com/c/6/a/c6ad63f4de5ccd9961e6450a642ddb8d46c03977.png)
such that 
where 
denotes the (non-negative) amount he bid for object .s are given beforehand, can we compute the set of all mixed strategy Nash equilibria?
by 
, 
for every positive integer 
. Prove that 
leave distinct remainders when divided by 
.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
for which there exists positive integers 
, 
, and 
satisfying![\[
2a^n+3b^n=4c^n.
\]](http://latex.artofproblemsolving.com/0/2/9/0295405b01494edc05ce9867b2ec48a4e069bf26.png)
and 
be two sequences of integers such that 
and 
for all 
. Suppose that 
divides 
for an infinity of integers . Prove that there exist an integer such that 
for all
are coprime. Consider the sequence 
given by 
. Suppose that there exist positive integers 
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than .
be an integer and let 
be distinct subsets of 
. Show that there exists 
such that the n subsets 
are also disjoint. be a positive integer, and consider the equation:
Answer the following questions. You may assume that 
is known.
Determine the number of real solutions of equation 
for each .
For , let 
be the segond largest real solution of .
Find 
such that 
Find 
, where 
is defined as
satisfies 
. If 
, find the maximum possible value of 
.
such that for all irrational numbers 
and 
,
be an integral domain and ![$A[X]$](http://latex.artofproblemsolving.com/8/f/c/8fc58a509a695c9718230e43a63ed145a6ce2835.png)
be its associated ring of polynomials. For every integer 
we define the map ![$\varphi_n : A[X] \to A[X],$](http://latex.artofproblemsolving.com/4/6/6/466be576a494f1b7fc51e8a0dacdc953fcbac5a3.png)
and we assume that the set ![$$M= \Big\{ n \in \mathbb{Z}_{\ge 2} : \varphi_n \mathrm{~is~an~endomorphism~of~the~ring~} A[X] \Big\}$$](http://latex.artofproblemsolving.com/b/b/8/bb8acb92661d9f087b34ef4abca876d46317b463.png)
is nonempty.
such that 
Then
and so
for all 
In particular, 
implies that the characteristic of is nonzero. Let be the characteristic of 
Now, 
gives
But it is a well-known fact (see here for a proof) that 
if and only if 
for some integer 
and some prime number 
in which case 
So 
hence 
and 
for some integer 
then
be an infinite domain. Let be an integer and suppose that the map 
defined by 
is a ring homomorphism. Then 
and so 
implying that the characteristic of is a prime number, say 
So contains a copy of 
Now consider the polynomial ![$$g(x):=(1+x)^n-1-x^n=\sum_{k=1}^{n-1}\binom{n}{k}x^k \in \mathbb{Z}_p[x].$$](http://latex.artofproblemsolving.com/d/2/2/d228d64a4e1ca1ed1ca8ffed32f43c8df051dcd1.png)
Then for any 
and so every element of is a root of 
Thus 
because is not finite. So 
for all positive integers 
and hence, by the link in my previous post, for some positive integer 
being a domain is not needed. is a prime , then is in the desired form. are in due to freshman's dream.
, let 
be the largest power of dividing so 
is relatively prime with . Considering 
, we get 
, as desired. 
, from we deduce that the characteristic of is nonzero, say 
. Taking reduction of ![$A[x]$](http://latex.artofproblemsolving.com/4/d/5/4d57b2e7466f9447d4489dc8ef949681ab435b45.png)
mod for a prime factor 
yields that all elements in are a power of . Hence has only one prime factor so it is a power of . Suppose 
for some positive integer . Since 
has -adic valuation equal to , it is nonzero if 
. This is a contradiction by considering 
. Thus 
and the Claim finishes. 
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
s
.
where 
