be the set of all positive integers. Find all functions 
such that 
divides 
for all positive integers 
and 
.
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each 
or 
.
are given. Losyash likes some of the cells of the 
checkerboard. In addition, he is interested in any checkered rectangle with a perimeter of 
,
,
.
is the midpoint of side 
of an acute—angled triangle 
.
is symmetric to the orthocenter 
inside triangle 
.
.
and 
Prove that
be real positive number satisfy that: 
.
,
integers and 
satisfy: 
are divisible by k. Prove that:
is divisible by k
such that 
and
.
and 
Prove that![$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$](http://latex.artofproblemsolving.com/5/6/1/561a990e5885863c77d454ffa3f1973e1bbf60b4.png)
![$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$](http://latex.artofproblemsolving.com/4/b/1/4b14911672688250c909571d788aaf10724462eb.png)
Where 
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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 
.
,
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 
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 
