# 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 $2$?

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$\forall x\in\mathbf{R}$,$\frac{p(x)-p(x-2)}{2}=2x-7$I don't think a polynomial with a degree higher than 2 will have such property. – Luqing Ye Dec 5 '12 at 18:37
Because of this formula: $\frac{(x+h)^n-x^n}{h}=(x+h)^{n-1}+(x+h)^{n-2}x+\cdots+(x+h)x^{n-2}+x^{n-1}$ – Luqing Ye Dec 5 '12 at 18:57

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

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I was thinking about that $+14$ above and could guess that was just for misleading. – Babak S. Dec 5 '12 at 18:39
@Amr That's a good answer. Thanks! – RicardoCruz Dec 5 '12 at 18:52

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 $q=r$.

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@Clive_Newstead I liked your answer too, because we don't need use derivatives. Thanks! – RicardoCruz Dec 5 '12 at 18:59
@Clive_Newstead : Nice approach – Phani Raj Feb 14 '13 at 19:12

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 $(x-2)^n-x^n$

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I think that u need to show why $p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$ will not work even if $a_{n-1},a_{n-2},...$ are non zero when n>2 – Amr Dec 5 '12 at 18:54