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of degree 
having 
distinct real roots (
), such that for every root 
: 
.
defined on all real numbers and taking real values such that ![\[f(f(y)) + f(x - y) = f(xf(y) - x),\]](http://latex.artofproblemsolving.com/f/5/2/f528fc4047e21247663d4993913eadf346315be7.png)
for all real numbers 
. Prove that there does not exist a pair of rational numbers 
such that 
.
![$P\in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/7/4/c/74cb2247069abb8a22666e87fcbd69a55a015004.png)
for which there exists 
such that for all 
we have: ![\[P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).\]](http://latex.artofproblemsolving.com/9/0/9/90954cdeade9d13149e1aaa03df2c87dff157041.png)
of 
such that the value of 
is fixed for all 
, and its value is an integer multiple of 
.
is fixed for all 
, and it follows that for any given value of there is at most one possible permutation:
such that 
. This has to be a multiple of so it is enough to talk about remainders of 
and .
combinations. is the number of permutation where is multiple of . fixed. i understand that we should have 
where is fixed.
where 
{
}.
if 
divide 
, if not 
or 
we have (one choice, 
)
we should have 
(one choice, 
) because it's the only way to have 
of the permutation are 
follows by 
of the permutation are 
follows by 
divide 
