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.
Where 
are positive integers and 
is a non-square positive integer. If 
denotes the n-th set of solution to the equation with 
Then, prove that 
, 
, 
be positive real numbers such that 
. Prove that ![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
be an integer. Prove that there is a set of 
points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.
be a triangle with circumcircle 
. Arbitrary points 
on 
respectively. Circumcircle 
of triangle 
intersects at 
. 
intersects 
at 
. 
cuts and at 
respectively. Construct parallelogram 
. Prove that 
are collinear.
be the unit circle in the complex plane. Let 
be the map given by 
. We define 
and 
for 
. The smallest positive integer such that 
is called the period of 
. Determine the total number of points in 
of period 
.
)
have real roots and real coefficients with 
. What is the maximum value of 
?, all the roots added will give 
. I also learned that if every single root, multiplies every single root, it will give 
. I also learned that if all the roots are multiplied, it will give 
. Is this right? And is there any purpose for these equations?
. Find the minimum and maximum values of 
+ 
+ 
and 
Show that
Let 
Show that
with 
.![\[
\frac{a^5}{b^2 + 2c^3} + \frac{2b^5}{3c + a^6} + \frac{c^7}{a + b^4} \geq 2.
\]](http://latex.artofproblemsolving.com/f/f/5/ff5043705025592aeefc55a764a8b7170e13ff4b.png)
be real numbers such that 
.
,
.
have roots 
.
(I won't be proving this here). This new information allows us to utilize the mean inequality chain as follows:
Similarly, applying AM-HM yields 
.
.
into the expression, you might notice things simplify quite nicely 
.
,