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.
. At a maths camp, there are 
children, each of whom selects
tables, with children tables, with children
and 
. Prove that
Let and 
. Prove that
and 
. Prove that
Where 
and 
. Prove that
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
. Let 
be the circle with diameter 
. Let 
be the intersection of 
and 
. Let 
be the orthocenter of . Let 
be the intersection of 
and . Let 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
. Let 
be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
![\[ f\left( f(x) y + \frac xy \right) = xyf(x^2+y^2). \]](http://latex.artofproblemsolving.com/b/6/5/b656324b6499f7457b7b9eccbf5eaa01b6ad6e59.png)
and 
be positive real numbers such that 
. Prove that
, where 
and 
.
satisfies the following property:![\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]](http://latex.artofproblemsolving.com/3/b/d/3bd9fbf82b0b1d98a3c68745bb3f159cd9e2529a.png)
and for all 
.
, let 
denote the number of elements in the set .
be the set of all functions![\[ f : \{1, 2, \ldots, n\} \to \{1, 2, \ldots, k\} \quad \text{with } n \geq 3,\; k \geq 2 \]](http://latex.artofproblemsolving.com/b/1/b/b1bf08934378882b9cd0b7d983f83b15453acca9.png)
satisfying the condition:![\[ f(i) \ne f(i+1) \quad \text{for every } i,\; 1 \leq i \leq n-1. \]](http://latex.artofproblemsolving.com/f/d/1/fd11b44d7fc482b5c6018e19161f6e30b7835c2d.png)
Show that![\[ |F| = k(k-1)^{n-1}. \]](http://latex.artofproblemsolving.com/a/8/8/a88a07984bad88a854de4dad8c24424aaad38365.png)
denote the number of functions in satisfying 
.
, show that![\[ c(n, k) = k(k-1)^{n-1} - c(n-1, k). \]](http://latex.artofproblemsolving.com/8/7/2/87285a27401b9853e72b45249196954082d9ec30.png)
,![\[ c(n, k) = (k-1)^n - (-1)^n(k-1). \]](http://latex.artofproblemsolving.com/2/f/c/2fcbcb385fcc1b3956232dadfd7ec4a31f4d71d6.png)
and 
such that 
is divisible by 
?
to deduce all of them are not divisible by 
denote the k-th symmetric sum and 
that 
and by newton sums we have 
and we consider the polynomial
which in 
is 
and the roots are 
so we get 
but 
and we are done
we get 
so as 
we have 
as desired.
, consider the polynomial 
, clearly 
and ![$P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/0/5/a05997bd7ba17001c62bc17ae35c76dd78e24789.png)
. And therefore we can let 
where all 
's are integers, so by taking newton sums modulo we have 
which gives that 
therefore by putting 
in this we get that 
for any 
which proves that all 's are congrent modulo and therefore we are done 
.
. Suppose 
of the 's are nonzero and WLOG they are 
. Let 
be a generator of 
. Let 
where 
is an integer for 
. Then the condition becomes 
for all 
. It follows that the elements of are roots of ![$P(x):=x^{\alpha_1}+\cdots+x^{\alpha_r}\in\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/8/b/0/8b0a5b5937b0aff17391fefd8df1603fb6c521a8.png)
. Thus 
is divisible by![\[
\prod_{a\in\mathbb{F}_p^\times}(x-a)=x^{p-1}-1.
\]](http://latex.artofproblemsolving.com/c/9/d/c9d1ccf7462e298113392f4798239e5f502ae2f4.png)
Since 
if is nonzero, 
and thus either 
in which case we are done, or 
and all the 
's are equal, as desired. 
Prove that
Prove that
be a prime number and 
be integers for which 
is divisible by 
.
is divisible by 
)
denote the 
-
,
.
(interpreted in ![$\mathbb F_p[t]$](http://latex.artofproblemsolving.com/d/1/f/d1fa992d9fe693e162a54b2121d646af3dd9f0e2.png)
)
,
.
.
But polynomials in 
modulo