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 non-negative real numbers, such that![\[a+b+c+d = a^2+b^2+c^2+d^2.\]](http://latex.artofproblemsolving.com/f/7/a/f7a1c451b34f277f8df35c8b1dca6510d3fc4c19.png)
Prove that![\[ab+bc+ca+da+db+dc \geqslant a^2b^2+b^2c^2+c^2a^2+d^2a^2+d^2b^2+d^2c^2.\]](http://latex.artofproblemsolving.com/c/e/0/ce016ac026a0cc3b78767629005ebf82916444b5.png)
(2) Let 
be non-negative real numbers, such that![\[\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} a_i^2.\]](http://latex.artofproblemsolving.com/d/5/3/d53b5ea7706a909bcf528a05b92690b72de5217d.png)
Prove that![\[\left( \sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2 \geqslant \left( \sum_{i=1}^{n} a_i^2 \right)^2 - \sum_{i=1}^{n} a_i^4 .\]](http://latex.artofproblemsolving.com/e/0/1/e0137b4bfa9bbe2371031f4c47df381a0ea38813.png)
Prove that:
When does the equality holds ?
have the Dirichlet propriety, if 
, 
.
with 
subset of 
and have the Dirichlet propriety.
?
?
intersect at 
. An arbitrary line through 
meets at 
respectively. The points 
are chosen on respectively so that 
. Suppose that 
meets 
at 
, and 
meets 
at 
. Prove that 
are collinear.
be non-negative real numbers such that 
. Prove the inequality![\[
6xyz \leq x(1 - x) + y(1 - y) + z(1 - z),
\]](http://latex.artofproblemsolving.com/f/5/b/f5b208ea6439994cb40f6b7f0806c216d9868e86.png)
and determine when equality holds.
, and 
be positive integers such that
If 
, prove that the polynomial
has no positive roots.
, such that: 
Prove that:
When does the equality hold ?
unit squares where 
. Two cells are called neighbors if they share a horizontal or vertical border. In the beginning, all cells together contain 
tokens. Each cell may contain one or several tokens or none. In each turn, choose one of the cells that contains at least one token for each of its neighbors and move one of those to each of its neighbors. The game ends if no such cell exists. such that the game does not end for any starting configuration and choice of cells during the game. such that the game ends for any starting configuration and choice of cells during the game.![$[ n ]$](http://latex.artofproblemsolving.com/4/c/a/4caca647174f974de170d214409efd015799f144.png)
be set of 
. Consider 
be 
subset of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
such that ![$S_1 \cup S_2 \cup \cdots \cup S_{n-1} = [n]$](http://latex.artofproblemsolving.com/b/3/e/b3e4063c64e12ae5c645732475ffde30e26cd011.png)
and 
for 
. Prove that there exist set ![$T \subset [n]$](http://latex.artofproblemsolving.com/d/a/b/dab13038bebce879b3ad3cb99320c0906964384b.png)
such that 
for all .
the points 
and 
are the feet of the altitudes through and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
be the circumcircle of triangle with 
. Let 
be a point on 
and 
be a point on the circle , such that 
. (The points 
and lie on different sides of the line 
). The line 
intersects for the second time in point . Show that 
.![\[
\begin{cases}
b^2 + 1 = ac, \\
c^2 + 1 = bd,
\end{cases}
\qquad (1)
\]](http://latex.artofproblemsolving.com/f/0/b/f0b02d1f56bfc12a2f72618b6a12822b1ad5fe0b.png)
where 
are positive integers.
is a solution of (1), then![\[
a = 3b - c, \quad d = 3c - b.
\]](http://latex.artofproblemsolving.com/9/b/d/9bd0e3d39592062f1cc5620b5d13729843d0c805.png)
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
be a non-constant monic polynomial with non-negative integer coefficients such that 
. Let 
be a sequence such that 
and 
. Prove that for every 
, there exists a prime 
such that 
but 
.
and integers 
, 
. Since 
, we get 
Suppose that for some 
, and each dividing 
, there exists 
such that 
. Then from the claim we get 
But we have 
which is a contradiction., there exists a prime such that but .
.
is injective modulo 
for any 
prime (e.g., see my solution to a 1999 Balkan problem).
such that 
. Take such a pair 
with the smallest coordinate sum and assume 
. Then, 
forces 
, contradicting with minimality of . So, 
. But then, there is a 
such that 
, forcing 
.
there exists no multiple of in the sequence. Now, look at the minimum index 
for which there exists 
such that 
. Clearly 
since then,![\[c=a_1 \equiv a_j=a_{j-1}^3+c \pmod{p}\]](http://latex.artofproblemsolving.com/6/2/2/62204ea0babff326c61bd41a039754344b409f53.png)
implying that 
which is a clear contradiction. But now,![\[0 \equiv a_j-a_i = (a_{j-1}^3 + c)-(a_{i-1}^3+c) = (a_{j-1}-a_{i-1})(a_{j-1}^2+a_{j-1}a_{i-1}+a_{i-1}^2) \pmod{p}\]](http://latex.artofproblemsolving.com/8/c/5/8c55158e6931618316ee6abf2cbc4575b47829df.png)
Hence, since 
(as this contradicts minimality) we have 
. However, for any prime 
(i.e a prime such that 
is not a QR 
) we have,
which is a contradiction. Hence, it is impossible that the given sequence has only finitely many primes which proves the desired result.
is finite
,
.
?
and 
for any prime 
Indeed, we proceed using an induction.
,
for some 
.
where there are 
to finish the induction.
for any prime 
:
for some 
to finish the induction.
be the set of prime divisors of 
.
is infinite, as required. 