Polynomial Problem, $P(x+1)P(x-1)=P(x^2+1)$ $P(x)$ is a real polynomial such that $P(x+1)P(x-1)=P(x^2+1)$. Find $P(x)$.
I have no idea how to start on this problem. The only things I could do was finding things like $P(2)P(0)=P(2)$ or differentiating $P(x+1)P(x-1)=P(x^2+1)$ and substituting some values for $x$.
 A: Let $R$ be the set of roots of $P$, and $S = \{z-1: z \in R\}$.  If $s \in S$,
$s+1$ is a root of $P$, so $s^2 + 1$ is a root of $P$, i.e. $S$ is invariant under $s \to s^2$.  Any finite set invariant under $s \to s^2$ must be contained in the union of the unit circle and $\{0\}$ (call this $T$). 
But also if $t-1$ is a root of $P$, $P(t^2+1) = P(t+1)P(t-1) = 0$ so $t^2+1$ is a root of $P$.  That is, if $s \in S$, $(s+2)^2 \in S \subseteq T$, implying $s+2 \in T$.  The only member of $T$ such that $s+2$ is also in $T$ is $-1$.  But $(-1)^2 = 1$, which can't be in $S$.  So there are no roots, and $P$ is a constant.  And $P^2 = P$ implies that the constant must be $0$ or $1$.
EDIT: OK, where to start in explaining this?  I hope you know something about 
complex numbers.
First of all, you have the 
Fundamental Theorem of Algebra which says that any polynomial $P$ can be written as $P(x) = a (x - r_1) \ldots (x - r_n)$ where $a$ and $r_1, \ldots, r_n$ are complex numbers: $r_1, \ldots, r_n$ are the roots of the polynomial.
The first step is to determine what the possible roots of your polynomial are, and after that you can try to determine $a$.
$R$ is the set of all
roots of the polynomial $P$, and $S$ is the set that you get by subtracting $1$ from every member of $S$.  If $r$ is a root of the polynomial $P$, and $s = r - 1$, then 
$P(s+1) = 0$ so $P(s^2+1) = P(s-1) P(s+1) = 0$. 
This says that $s^2 + 1$ is a root of $P$, so $s^2$ is in $S$.  So if you take any member of $S$ and square it, you again have a member of $S$.  
Now if you have any complex number $s$ and keep squaring it, what happens?
If its absolute value $|s|$ is greater than $1$, the absolute value increases each time you square it, and so you get an infinite sequence of numbers with larger and larger absolute value.  But there are only a finite number of roots of the polynomial $P$, so these numbers can't all be in $S$.  That says there can't be any members of $S$ with absolute value greater than $1$.  Similarly,
if you square a number of absolute value less than $1$ but not $0$, the absolute value gets smaller each time.  Again, we can't have that because 
there are only a finite number of roots.  So we conclude that every member of
$S$ has absolute value either $0$  or $1$.  Geometrically, these candidates for $S$ form the circle of radius $1$ centred at $0$, together with $0$ itself.
Now, what about a number of the form $t = r + 1$ where $r$ is a root of $P$?
The equation says $P(t+1)P(t-1) = P(t^2+1) $, but $P(t-1) = 0$, so $P(t^2+1) = 0$, i.e. $t^2 + 1$ is a root
of $P$, and $t^2$ is in $S$.  Again, the absolute value of $t^2$ must be $0$ or $1$, but $|t^2| = |t|^2$ so the absolute value of $t$ itself is $0$ or $1$.
Thus if you take $s = r-1$, which is either $0$ or a point on the circle,
and add $2$, you get $t = r+1$, which is again either $0$ or a point on the circle.  The only way you can do that is if $s = -1$ so $t = 1$.  That is, the only possible member of $S$ is $-1$.  But then $(-1)^2 = 1$ would also have to be in $S$, and that is impossible.
I hope that helps.  If there are any parts that you still don't understand, please ask.   
