Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $2$?

share|cite|improve this question
$\forall x\in\mathbf{R}$,$\frac{p(x)-p(x-2)}{2}=2x-7$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
up vote 7 down vote accepted

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

share|cite|improve this answer
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 $q=r$.

share|cite|improve this answer
@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 $(x-2)^n-x^n$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.