Prove that
Let 
Prove that
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
Prove that
be vertices of regular pentagon inscribed in a circle whose radius is 
and center is at 
.
be an integer. Find all positive real solutions to the following system of 
equations:
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
be the reflection of 
.
intersects line 
and 
.
.
such that 
is an integer for every positive integer 
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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
in 
is actually 
each element has order 
is 
,
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 
relatively prime to 
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 
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 
