for which the inequality ![\[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]](http://latex.artofproblemsolving.com/4/d/b/4db42ebabc7309bba9629a9a7125be43abf113f0.png)
holds for all real numbers 
,
,
,
.
and 
are given. Let 
be the midpoint of the segment joining their centers, 
,
be arbitrary points on 
.
.
,
be the circumcenter and the incenter of an acute-angled scalene triangle 
;
,
,
be the touching points of its excircle with the side 
and the extensions of 
,
respectively. Prove that if the orthocenter of the triangle 
lies on the circumcircle of 
.
be a point inside a quadrilateral 
such that 
.
,
,
are isogonally conjugated to 
,
,
,
concur.
can be written in the form![\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]](http://latex.artofproblemsolving.com/9/e/9/9e96c07ea3b0750af8bed344df6e8a86fcf0d772.png)
where 
and 
are positive integers. Find the value of 
.
be an acute triangle with 
,
.
be the altitude from vertex 
.
and 
be the feet of the perpendiculars from 
to the lines 
and 
,
intersect the circle 
and 
such that 
.
be the midpoint of segment 
intersect line 
at point 
.
.
is parallel to line 
.
be the orthocenter of triangle 
,
be the midpoint of segment 
is the center (HBC). Let point 
lie on line 
such that ray 
is the angle bisector of 
.
intersect at a point lying on circle 
(i.e., the square on 
has 
,
,
to the centers of the opposite squares are concurrent.
be positive integers. Maximize 
if 
.
,
,
.
where 
is a natural number.What is the remainder when 
Y by Adventure10, asdf334, donotoven, mathematicsy, TFIRSTMGMEDALIST, megarnie, adityaguharoy, ImSh95, Rounak_iitr, Mango247, MS_asdfgzxcvb, and 1 other user
Determine all positive integers 
that satisfy the following condition: for all 
and 
relatively prime to 
we have ![\[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]](//latex.artofproblemsolving.com/e/8/5/e85c5b249ac8e75b45a525b8786414798a4921fe.png)
that satisfy the following condition: for all 
and 
relatively prime to 
we have ![\[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]](http://latex.artofproblemsolving.com/e/8/5/e85c5b249ac8e75b45a525b8786414798a4921fe.png)
This post has been edited 1 time. Last edited by djmathman, Oct 3, 2016, 3:32 AM
Reason: adjusted formatting
Reason: adjusted formatting
is actually 
each element has order 
.
,
and 
.
,
is the smallest prime factor of 
: if 
then this might not work; Anyway, we know that 
must be a power of 
,
odd primes of the form 
.
and 
.
then we take 
.
doesn't divide 
.
,
.
that we have to check is 
,
for all 
?
for 
to get 
values, get that 
,
results in 
,
,
results in 
,
also works but 
does not.
,
,
results in 
,
,
,
does not work, so it has to divide 
,
for large 
is the nth prime. This can be proved by induction because it is true for 
and if 
,
,
for any 
.
where 
thus from the condition 
Hence 
so 
.
.
divides 
where 
are positive integers. Let 
be a primitive root of 
.
where
Clearly the order of 
in 
is 
.
which must be distinct and multiply to 
.
,
which implies 
.
where 
,
or 
or it can be both.
are the only primes satisfying this.
.
holds,
that work are those satisfying 
.
and 
,
.
,
,
,
,
,
,
,
,
.
and 
,
iff 
.
which we can check.
,
,
,
,
and 
for all a,b 
,
,
which is not 
fails since 
,
.
mod 
,
for all 
or 
.
and doesn't work when 
.
Therefore, for each prime power divisor of 
if and only if 
Therefore, 
Testing gives 
and 
so the answers are just the divisors of 
dividing 
must be true if 
,
are relatively prime to 
that there exist 
that are relatively prime to 
or 
.
for some 
,
we get 
.
.
and 
and 
then 
By Chinese Remainder Theorem, 
where 
We can say 
by using 
We can easily verify the solutions 
and 
for every prime 
.
,
(mod n) 
works, so the answer is 
