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 positive real numbers . Prove that
satisfying
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
be positive reals with 
. Then show that![\[\sum_{i=1}^n \frac{x_i}{2-x_i} \ge \frac{n}{2n-1}\]](http://latex.artofproblemsolving.com/f/6/5/f65f6fce4465125dfacf002f0659c898df7ae150.png)
and 
are the incenter and the 
-excenter of triangle 
, respectively.
be the midpoint of arc 
on the circumcircle, and 
be the foot of the to 
. The line 
intersects the circumcircle again at 
. For
on the circumcircle of triangle , let 
intersect at 
. Prove
are concyclic.
squares, we cover the area with several S-Tetrominos (=Z-Tetrominos) along the square so that in every square, there are two or fewer tiles covering that (tiles can be overlap). Find the maximum possible number of squares covered by at least one tile.
satisfies
for any non-negative integers 
and 
. Determine the number of possible sets of integers 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
, centroid 
and orthocenter 
. Prove that 
is constant.
of positive integers, such that every sum
(with 
) is a composite number, but:
for all 
; for all and 
for all 
?
denotes the greatest common divisor of 
, 
.
of positive integers, such that every sum(with ) is a composite number, but: for all ; for all and for all ? denotes the greatest common divisor of , .
, there exist sequence 
of positive integers such that 
is composite number for all 
and 
for all 
and 
for all 
, we choose sequence 
, then we choose 
where 
is prime number larger than 
and 
is a positive integer such that 
where 
is a prime divisor of 
for all 
will give us for all , 
and 
for all 
, and another bunch of odd primes (pairwise distinct and also distinct from ), say, 
. Now by virtue of CRT, choose a number 
such that:
Now set 
for 
. We claim that this works. Indeed, since the 
's are in an arithmetic progression, we have 
Now 
is always an integer, because by construction, and hence all 's are odd. For 
, the above sum is obviously composite, and for 
, the above sum reduces to 
, which is composite because 
.
, which can't be 
since we've been careful to set 
. The case 
is similar, so we win. 
and so on
then you have that sum is 
which is obviously incorrect since 
is not a composite number.
, then 
, 
and 
. Then inductively construct the 's for 
as follows.
, 
for 
and 
. This can be done by CRT as we can take sufficiently large 
's and 
's to ensure that they are pairwise coprime and also coprime to 
.
consecutive terms of a arithmetic progression whose each term is odd composite number is a composite number.
where n is odd.Clearly the sum ![$ n[x+\frac{n-1}{2}d]$](http://latex.artofproblemsolving.com/d/b/8/db8452ec6c74f166c70507213019065eb9b081c2.png)
is composite. there is a arithmetic progression with terms such that the following holds:
.Consider the AP,
.
and common difference 
and first terms. devides GCD of 2 terms,
. devides at least one of 
or but in both cases,as 
,
.The 4th line follows from the fact that 
.
by the above 2 claims both (a),(b) holds .
for all 
, where 
.
and similarly 
.
for all , we get 
and since 
we are done!
. (Goal being )
just take 
or anything similar.
we choose primes 
such that they are all distinct and coprime to any subarray sum we have so far. Now we construct 
using 
such that:![\[ a_{n+1} \equiv -\sum_{i=j}^n a_i \mod p_{n,j}^2\]](http://latex.artofproblemsolving.com/a/1/0/a1084b5f9940fd8157deedb9a72c6b71f70534e2.png)
for all 
and![\[ a_{n+1} \equiv 1 \mod a_na_{n-1} \]](http://latex.artofproblemsolving.com/6/7/5/675b5b20c0d78964e79ebd1bf85416a31b9a40bd.png)
This is valid, since each of the squares are coprime to all elements, including 
. Since each new subarray containing is divisible by some square, they will not be prime. Additionally, is coprime to both 
and 
, so the construction is valid for 
and by induction for all , including 2016
:
