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.
are of the form 
, where 
, 2, 3, . For each 
, let 
be the greatest common divisor of 
and 
. Find the maximum value of as ranges through the positive integers.
be non-negative real numbers such that 
Prove that
be positive real numbers such that 
and 
, prove that
and 
outside triangle 
such that 
and 
. Connect 
and 
, which intersect at point 
. Let 
intersect 
at point 
. Prove that 
.
for which each cell of 
table can be filled with one of the letters 
and in such a way that:
, one third are 
and one third are ; and [/*], one third are and one third are .[/*] table are each labelled 
to in a natural order. Thus each cell corresponds to a pair of positive integer 
with 
. For 
, the table has 
diagonals of two types. A diagonal of first type consists all cells for which 
is a constant, and the diagonal of this second type consists all cells for which 
is constant.
,![\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]](http://latex.artofproblemsolving.com/b/5/d/b5d49e25bd8d8a211b72372ff967558e1a8365ce.png)
Find the smallest positive integer such that 
is divisible by 200.
. Prove that
Let 
. Prove that![$$ \frac{1}{2}\geq \frac{a}{a^2+3 }+ \frac{b}{b^2+3} \geq \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$](http://latex.artofproblemsolving.com/f/9/1/f9119e071c12778037616c624c7b9ca75356e8e3.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+3 }+ \frac{b}{b^3+3} \geq \frac{1}{2\sqrt[3]{9}}$$](http://latex.artofproblemsolving.com/2/b/2/2b2d8a011229fe88a81cb75f9b018ad3a6245faa.png)
![$$ \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2} \geq \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$](http://latex.artofproblemsolving.com/5/f/6/5f673cd494b1909ca237db63e96f949c24b6f2ad.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2} \geq \frac{\sqrt[3]{3}}{5}$$](http://latex.artofproblemsolving.com/b/c/3/bc343a07c88dbd98ebffdfb78f311b7b82683d65.png)
.
Let 
.
and 
Prove that
Let 
Prove that
