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.
such that divides 
but 
does not divide 
.
and such that:![\[ 3n^2+3n+7=m^3 \]](http://latex.artofproblemsolving.com/6/f/8/6f84f81f857094b3ea432284a7881697167b2418.png)
be a point on the Neuberg cubic. Show that as varies, the Nine Point Circle of the antipedal triangle of goes through a fixed point.
for which there exists a positive integer 
with the property that is divisible by and 
is divisible by 
.
be the circumcircle of a non-isosceles triangle 
and 
be a tangent line to (
). Let 
be midpoint of 
and 
. Prove that 
and 
be coprime positive integers such that:![\[ \dfrac{p}{q}=1-\frac12+\frac13-\frac14 \cdots -\dfrac{1}{1334}+\dfrac{1}{1335} \]](http://latex.artofproblemsolving.com/b/d/3/bd374388fe6b84fbcd66e23a7956f2fbeebe4c28.png)
is divisible by 2003.
and all positive integers 
and 
satisfying 
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)
is interesting if 2018 divides 
(the number of positive divisors of ). Determine all positive integers 
such that there exists an infinite arithmetic progression with common difference whose terms are all interesting.
be an odd prime number and a sequence 
satisfy 
and 
Prove that always exists positive integer satisfying for all positive integers , then 
.
is "not congruence"
, the altitude at 
, the bisector of 
and the median at 
meet at a common point. Prove (or disprove?) that the triangle is equilateral.
and to be less than 
. Using the obvious relations for the segments determined on the sides, and Ceva's theorem, we need have 
, so 
.
, 
, 
are fulfilled for any values 
, so such a triangle exists for any values ; only when 
it follows 
, thus an equilateral triangle.
and to be less than . Using the obvious relations for the segments determined on the sides, and Ceva's theorem, we need have , so ., , are fulfilled for any values , so such a triangle exists for any values ; only when it follows , thus an equilateral triangle.
and then we have 
, and having 
and 
, whence mine relation 
, and not yours 
. Thus the conditions look just as I described them.
, and having and , whence mine relation , and not yours . Thus the conditions look just as I described them.
and 
.
,
.
that satisfies:
