[/quote]
,
,
that makes the following equation have no integer root 
.
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
,
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
and 
: Prove that :
with 
.![\[
(x^{2}-yz)(y^{2}-zx)(z^{2}-xy).
\]](http://latex.artofproblemsolving.com/a/8/7/a87fae27584012a867bc0a90190c0317f9f7ab07.png)
and 
such that 
.
is a benefactor of 
because 
.
benefactors.
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
such that for any positive real numbers 
,![\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]](http://latex.artofproblemsolving.com/9/1/c/91c8a4af2bd2f3107c8921d2e06f8a09dedd0164.png)
.
meets 
at points 
,
is 
.
intersects line 
at 
,
intersects line 
,
intersect at 
.
is tangent to 
.
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
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 
,
mod 
mod 
or 
.
.
.
.
(except 
as it is excluded). Clearly all these numbers work, so we are done. 
,
for contradictions.
for all 
?
for 
relatively prime to 
to get 
values, get that 
,
results in 
,
results in 
,
also works but 
,
,
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 
,
.
,
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 
