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.
interesting if 
is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
+1 w
such that 
holds for all real numbers 
and 
.
be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
be a triangle with 
. Let 
be the midpoint of 
, and let 
be a point on segment such that 
. Let 
be the point of intersection, different from 
, of the circumcircle of triangle 
and line 
. Define 
and 
as the points of intersection of line 
with 
and 
, respectively. Prove that is the midpoint of 
.
have orthocenter and altitude 
with 
on side 
. Let be the midpoint of side , and let 
be the reflection of over . Let 
be a point on line 
such that lines 
and are parallel, and let the circumcircles of 
and 
meet again at 
. Prove that 
. students and their test results on 
different subjects, we say that student 
dominates student 
if and only if outperforms on all subjects. Assume that no two of them have the same score on the same subject, find the probability that there exists a pair of domination in class.
of positive integers that satisfy the equation![\[
x + xy + xyz = 31.
\]](http://latex.artofproblemsolving.com/a/7/8/a78d1e9d66831f11e7de08c109855e7727ec8535.png)
be real numbers such that 
, and![\[
a + \frac{1}{a} + b + \frac{1}{b} + c + \frac{1}{c} + d + \frac{1}{d} = 0.
\]](http://latex.artofproblemsolving.com/a/9/0/a90c94d82e3650560635616dacaf392d300b8c91.png)
Prove that one of the numbers 
or 
is equal to 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
+1 w
for which there are infinitely many positive integers such that 
is not divisible by 
.
and 
then we know that modulo 
. This means that in base , in the addition of 
and 
we have at least 
carries, which implies by Kummer's theorem that 
so if all prime divisors of are not 
we know does not work. So if 
we have it doesn't work.
works for large enough (as in, 
) and setting 
works for 
, and that with a little bit more details we can make it work for the rest of the numbers.
such that 
,
which satisfies
