How to prove the uniqueness of a polynomial with such properties? Let $p(x)$ be a polynomial of degree $n$ with real coefficients such that $$p(x-2)=p(x)-4x+14$$ for every real number $x$ and $$p(0)=6.$$ How can we prove that $n$ is $2$ and not higher than $2$?
 A: By differentiating twice we get:
$$p^{(2)}(x-2)=p^{(2)}(x)$$
Thus, $p^{(2)}(x)$ is  the constant polynomial. Now it follows that the degree of $p(x)$ is 2
A: Amr's answer is much simpler, and was posted whilst I typed this out. I'll post this anyway because it's a different approach that doesn't require calculus (but is clunkier).
Restrict to even integer values of $x$: if $a_n = p(2n)$ for integer $n$, then rearranging gives
$$a_n - a_{n-1} = 8n - 14,\ \ a_0=6$$
so the sequence $(a_n)$ is one where the difference between any two terms is linear; hence $a_n$ is a quadratic polynomial in $n$.
So $p(2n)$ is a quadratic polynomial in $n$. This means that $p$ agrees with a quadratic polynomial at every even integer. But this means that it must be equal to this quadratic polynomial everywhere.
Why? Because if $q$ and $r$ are polynomials and $q(x)=r(x)$ for infinitely many values of $x$ then $q(x)-r(x)=0$ for infinitely many $x$, so $q(x)-r(x)$ has infinitely many roots, so it must be zero, and so $q=r$.
A: The solution of this problem is $p=x^2-5x+6$. It should be easy to prove that any formula $p=ax^n$ with $n>2$ will not work, by filling it in in the formula, and expanding $(x-2)^n-x^n$
