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.
find the number of ordered real triples 
such that![\[\begin{aligned}
\begin{cases}
x+y^2+z^2=a,\\
x^2+y+z^2=a,\\
x^2+y^2+z=a.
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/9/8/6/986593653bf128bf7b3d497abed78e1459134c5f.png)
+1 w
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of 
numbers each in such a way that the sums of the numbers in these groups are pairwise distinct.
+1 w
such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](http://latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where 
is greatest integer not greater than 
such that ![\[ \dfrac{a^2}{2ab^2-b^3+1} \]](http://latex.artofproblemsolving.com/0/d/2/0d2f10786d22af2ea94ebeed6b15e723f4042734.png)
is a positive integer.
?
be a triangle and let 
and 
be variable points on sides 
and 
, respectively, such that 
. Let 
denote the point on 
such that 
is parallel to 
, and let 
denote the point on 
such that 
is parallel to 
. Prove that as 
vary, the circumcircle of 
passes through a fixed point, other than 
.![\[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]](http://latex.artofproblemsolving.com/7/9/e/79e388eba20412905964dc2b4ebb491f3c79143b.png)
and the sums ![\[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]](http://latex.artofproblemsolving.com/3/c/4/3c44766edcc9e5269f0d3396386e41676cbdf42c.png)
For any pair define 
and 
, 
.
, 
such that the sequences 
and 
are unbounded while the sequence 
is bounded?
, 
?
, angle 
. Point 
lies on 
such that 
. Line parallel to 
passing trough intersects 
at 
. Prove that 
.
be a positive integer. Find, in terms of , all polynomials 
with integer coefficients such that for every integer , there exists an integer 
such that 
.
.
is defined as follows: 
and ![$ {a_n} = \sum \limits_ {k = 1} ^ {[\sqrt n]} {{a_ {n - {k ^ 2 }}}} $](http://latex.artofproblemsolving.com/6/c/c/6cc22be2953e8b1556f149510a6712116e6a1fe2.png)
for 
there are at least 
even numbers.![$ [x] $](http://latex.artofproblemsolving.com/3/a/7/3a7c3166538c5f6ffbf90df21809c8c413a82607.png)
is the largest integer not exceeding 
.)
for all 
. It is easy to see that 
counts the number of sequences of positive integers 
such that 
. For each 
, let 
denote the set of all such sequences.
as 
. Since 
is an involution, the number of fixed points of has the same parity as 
. Consider a fixed point of 
. If 
is even, say 
, then 
. There are such sequences. If is odd, say 
, then 
. Note 
must be even, so let 
. Then 
. For each 
, there are 
such sequences. Therefore, 
Suppose 
is not a perfect square (so each sequence in has at least two terms). Define 
as 
. We similarly want to count the number of fixed points of 
. A fixed point occurs when 
for some . Then 
. There are such sequences. Therefore, 
Putting the two equations together, whenever is not a perfect square, 
This evidently gives more than even numbers in 
. 
is even if and only if is not a perfect square. With significantly more work, the parity of 
can be characterized. I was not able to characterize the values of 
, so I will be really interested in a solution that gives those values too.
satisfy that 
.
,
for 
For each 
,
Observe that there are at most 
elements which are in at least one of the 
'
positive integers ![$n \in [1, 10^6]$](http://latex.artofproblemsolving.com/b/3/c/b3ce635a2d217a05b8d8e8cbc3690ba77ed91069.png)
such that 
is even, there must exist some 
even, 
This means, however, that 
is odd for 
from the definition of the 
)