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 odd integers such that 
and 
. Prove that if 
and 
for some integers 
and 
, then 
.
+1 w
such that 
Prove that
Where 
such that 
Prove that
parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on is not greater than 
?
be reals such that 
and 
Prove that
Let be reals such that 
and Prove that![$$ a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$$](http://latex.artofproblemsolving.com/6/2/2/62211222c148c86d65526d86dde603cb6598925f.png)
![$$ a^2+2b^2+c^2\geq 2\sqrt[3]{4} $$](http://latex.artofproblemsolving.com/3/8/3/383d38cc87c1b1ea1e17e10e4bb95bed79329786.png)
be the set of positive integers. Find all functions 
, defined on 
and taking values in , such that 
for every positive integer 
.
be a nonconstant polynomial of degree with rational coefficients which can not be presented as a product of two nonconstant polynomials with rational coefficients. Prove that the number of polynomials 
of degree less than with rational coefficients such that divides 
.
be positive real numbers so that 
. Prove that![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
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)
be a triangle. Suppose 
is on 
such that 
bisects 
. Suppose 
is on 
such that 
, and 
is on 
such that 
. If and 
intersect on 
, prove that 
.
then 
is cyclic, being the midpoint of its circumcircle arc 
Hence 
and 
gives 
But 
and similarly 
hence 
Combining 
and 
yields 
is cyclic. bisects , 
and 
. Then 
. So, we proved.
and :
. So we are done! 
and 
.
such that: 
.
such that 
then find minimal value of 
Where 
