AoPS Academy
be a positive integer. Find the number of permutations 
, 
, 
of the
, 
, 
, satisfying
.
positive integer the equation: 
has positive integer solutions.
be an equilateral triangle. Let 
be a point inside of it such that the square root of the distance of to one of the sides is equal to the sum of the square roots of the distances of to the other two sides. Find the geometric place of .
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective. is bijective if for every integer 
, there exists exactly one integer 
such that 
.
+2 w is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer on the board with 
, and on Bob's turn he must replace some even integer on the board with 
. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends. and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.
and 
is closest to which integer?
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
, we have![\[
2^{2^x} + 1 \mid 2^{2^y} - 1,
\]](http://latex.artofproblemsolving.com/a/5/c/a5c3b28c3cbaaae4acb9e6fd80f3e45f589adb26.png)
since![\[
2^{2^y} - 1 = (2^{2^{y-1}} - 1)(2^{2^{y-1}} + 1).
\]](http://latex.artofproblemsolving.com/9/6/9/96929693f428528a395ec0287e02435bec5467ab.png)
Therefore, 
, which means the set of prime divisors of numbers of the form 
is infinite. Then we can choose 
and an odd prime 
such that![\[
p \mid 2^{2^{n_0}} + 1 \quad \text{and} \quad p \nmid a.
\]](http://latex.artofproblemsolving.com/b/3/a/b3a5ef39bc64c958af45eae01bc5a17e301f902f.png)
, where 
is an odd number; we will introduce additional conditions on it later. Then:
, where 
is odd. Then, if is sufficiently large and 
, then 
. Then, by Fermat's Little Theorem:![\[
a^{k^k} \equiv 1 \mod p \Rightarrow b_k \equiv \frac{1 \cdot 1}{-1} + 1 \equiv 0 \mod p.
\]](http://latex.artofproblemsolving.com/6/4/c/64c67cf1f60032c66aa40959f4954332acdf6c6c.png)
, we havesince
divide 
?
![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]](http://latex.artofproblemsolving.com/2/d/2/2d248ce9756f0f09265dd538de3b16b22a944c46.png)
divides 
,
.
and because 
,
If 
then 
Note that 
,
and
so we are done. 
?
,