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such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
with integer coefficients, satisfying 
for all non-negative integers 
.
be a triangle. 
and 
are points, such that 
, 
= 
and 
( - inside 
and - oitside). Proove that line 
passes through the fix point.
such that 
is a prime, 
and 
. Also, the following polynomial is reducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
![$\mathbb{Z}[x]. $](http://latex.artofproblemsolving.com/1/e/4/1e4231365651da420e8645c1246758a92a0cae78.png)
The original wording sounds something more like this:
be a prime, and let 
and be two distinct positive integers less than 
. is odd and 
can be expressed as a product of two non-constant polynomials with integer coefficients. Determine all possible 
and ![$\mathbb{Z}_p[x]$](http://latex.artofproblemsolving.com/3/f/d/3fd6af0acc8b77146f2f801902f10d7e893644f4.png)
. Suppose 
.
and assume 
for nonconstant monic polynomials ![$g,h\in\mathbb Z[x]$](http://latex.artofproblemsolving.com/b/6/7/b67e4575735b3f78b526941d2c46c8ce1c9a8c05.png)
, and let 
be the image of 
by the natural map ![$\mathbb{Z}[x]\rightarrow \mathbb{Z}_p[x]$](http://latex.artofproblemsolving.com/7/3/9/739ad963a07248caee1ed36db696f020e94da721.png)
, respectively. We have
and since 
, our only choice is 
.![$A(x),B(x)\in\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/5/2/e/52eb2fe1e8bedb81296067b07d58fe59d30f0351.png)
such that 
. Rewrite the factorization as
, then we get 
We go back to . The equation becomes
The key observation is that since were monic, the degree of 
has to be at most 
. Thus 
and 
are precisely the 
and 
-degree parts of 
. This motivates us to divide cases.
. This implies 
, so we let 
. We have
but substituting 
gives 
, which is impossible for .
. This time 
for some 
. Plugging in 
in
yields 
This is again a contradiction for .
: 
all work.
: that no such 
exist.
.
is necessary, otherwise all triples in the form 
are solutions. We figured the problem was already getting too hard, so we resolved to odd . The problem for general is open.
),