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, such that 
and 
.
such that 
is prime, 
with 
and ?
with real coefficients obeying![\[P(x) P(x+1) = P(x^2 + x + 1)\]](http://latex.artofproblemsolving.com/b/e/f/beff3420fa8394f4f98c031d81eea6a5a3b8d28c.png)
for all real numbers 
.
such that 
divides 
for any integers 
and there exists a polynomial with integer coefficients, such that 
for all 
.
also divides 
.
such that 
but 
. Then, we can choose some 
such that 
. This implies 
, leading to 
, a contradiction.
must be a power of for all primes . Since 
for all 
, there exists some 
such that 
for infinitely many primes . It follows that 
for all , where is odd.
infinitely often along primes, let's justify the last step why 
for 
. Taking 
and keeping 
fixed, 
implies that 
. Using 
(the inverse exists as 
), we obtain![\[
p^k f(n) + 1\equiv (-1)^k \frac{f(n)}{n^k}+1\pmod{pn+1}\Rightarrow pn+1\mid \left|f(n)\pm n^k\right|.
\]](http://latex.artofproblemsolving.com/f/3/9/f392fe8bd086ec7abc433e8d35bb97f55e61f065.png)
By taking 
large, we are left with (as 
). A byproduct of this argument is that 
need be odd.
and![\[(x-y)[f(x)+f(y)]\leqslant f(x^2-y^2), \forall x,y \in \mathbb{R}\]](http://latex.artofproblemsolving.com/8/0/b/80bb1d7a992c9683a0c2a07b2febe6271f4e60f5.png)
such that 
