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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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    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}$2012-12-05

3 Answers 3

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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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    @Amr That's a good answer. Thanks!2012-12-05
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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 : Nice approach2013-02-14
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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>22012-12-05