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 integers.Define injective function 
from 
to itself such that 
.Prove that 
divides 
.
board that can be colored so that for each square in the board, at most one square adjacent to it is colored.
such that 
.
that divides 
for the positive integer
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
of positive integers such that 
in all of this cases :
of positive integers such that : 
3-Every prime number less than 
can be represented in the form 
by taking an Appropriate pair cannot be represented in the form where 
and are positive integers
. The number 
is the least prime number that cannot be represented as a sum or an difference of a power of and a power of . Prove that the number cannot be represented in any of the forms 
, 
, where and are positive integers
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
satisfying
of injective functions from the set of real numbers to itself satisfying
and 
. (We say a function 
is injective if 
for any distinct real numbers 
and 
.)
, find the smallest integer 
such that any graph with vertices and at least edges must contain at least two disjoint cycles (cycles with no common vertices).
with 
and 
Let 
such that 
and 
is bisector of 
Denote by 
the intersection of 
and 
, and by 
the intersection of 
and 
.
.
.
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
be real positive numbers such that: 
.
![\[
\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
\]](http://latex.artofproblemsolving.com/9/5/9/9591f7c842d1b6afcba11b15b3ea3f9e01bc937a.png)
![\[
2^a + 3^b + 5^c = n!
\]](http://latex.artofproblemsolving.com/b/c/6/bc6b0754dd874f646fcab124c91583b620db273c.png)
Then, we have 
Note that 
which implies that 
is a 
triangle with right angle at 
and 
angle at 
Thus, by the Angle Bisector Theorem, we have 
so 
Then, the line going through 
and 
has slope 
and 
so its equation is 
Similarly, we can derive that the equation of the line containing 
and 
is 
Therefore, the intersection of the lines 
and 
is 
Then, we can find that the slope of the line going through 
while the slope of the line going through 
Since these slopes are negative reciprocals, we can conclude that 
Since the slope of the line going through 
and this line goes through the origin, the equation of this line is 
Recall that 
so the slope of the line going through 
Its 
so it follows that the equation of this line is 
The intersection of this line and line 
Now, we simply use the distance formula to compute 
and 
so we are done.