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.
, where a > 0 and 
is differentiable such that 
Then which of the following is correct?
always exists but not necessarily zero. always exists and is equal to zero. may not exist. exists if 
is twice differentiable.
such that for all 
, ![\[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]](http://latex.artofproblemsolving.com/5/b/1/5b101ead6195155ccbd99b757016e8f73dac4411.png)
Proposed by Aritra Mondal
be positive real numbers such that 
. Prove that
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
, 
, 
checkered board is cut into several squares with integer side length. The size of a square is it's side length.
of integers, a pair 
with 
is called interesting if there exists a pair 
of integers with 
such that 
For each 
, find the largest possible number of interesting pairs in a sequence of length 
.
such that, for all 
,
cards on which the numbers 
, 
, 
, are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let 
be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer 
for which Bob can guarantee that is at least .
be positive real number such that 
Find the minimum value of 
mod 
{
} (see below)
mod {
}
mod {
}
for the LHS and 
or 
for the RHS in 
)
or 
, unlike the LHS. Thus also this option fails and no satisfies the given requirements.
and 
for positive integers 
and 
.
.
, so at least one of 
or 
has to be divisible by . and 
are divisible by , and 
., 
, and 
are pairwise relatively prime, we have three cases:
.
, which cannot be a perfect cube.
.
, which is impossible, since all perfect squares divisible by are divisible by 
.
.
, which is impossible, since all perfect squares divisible by are divisible by .
and 
cannot be perfect cubes simultaneously.
is clearly odd, as otherwise 
. Now, factorize this quantity as 
; denote by 
the gcd of the factors. Then, 
implies 
, thus 
. As is odd, 
. Now, if 
then 
for some 
for which 
, thus it cannot be a perfect cube. Hence, 
; leaving us with two cases.
. For 
, 
is not a cube. Let 
. Using 
we get 
, but![\[
\left(4a^3-1\right)^2<8a^6-6a^3+3<8a^6,
\]](http://latex.artofproblemsolving.com/8/e/f/8ef3f6e4ffff8e2d47b7acea79a5c3323194815c.png)
yielding a contradiction.
. In this case, 
. Note that 
, thus 
and 
(by Euler). Hence, 
, none of which is a perfect cube.
is a cube. The fact that
, yields 
since 
for all integers . Hence 
, which means
,
is not a cube. This contradiction completes the proof.
is a cube. The fact that, yields since for all integers . Hence , which means, is not a cube. This contradiction completes the proof.
So 
is also possible but then 
a contradiction.
prove inequality 
Prove that
Where 
is a perfect cube. Prove that 
is not a perfect cube.