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Find all polynomials $p$ with real coefficients such that $p(x+1)=p(x)+2x+1$.

I feel like in this question you let $x+1=x'$.

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  • $\begingroup$ What is $P(x)$? Do you mean $p(x+1) = p(x) + 2x + 1$? $\endgroup$ – Chris K Apr 14 '14 at 16:14
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    $\begingroup$ If so, one example would be $p(x) = x^2$. $\endgroup$ – Chris K Apr 14 '14 at 16:15
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Assume $p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0$ with $a_n\ne 0$. Show that $p(x+1)-p(x)=na_nx^{n-1}+(\text{lower terms})$. This tells you that $p$ must be of degree $n=2$ and that $a_2=1$, so $p(x)=x^2+a_1x+a_0$. Now you can comfortably compute $p(x+1)-p(x)$ explicitly and compare with $2x+1$.

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Just a different interpretation: For fixed $x\in [0,1)$ the equation gives a recursion or difference equation for $n\in\Bbb Z$:

$$p(x+n+1)-p(x+n)=2n+2x+1.$$

This has the constant sequences as homogeneous solutions and $An^2+Bn$ as type of a particular inhomogeneous solution. Inserting an comparing gives $$ A(2n+1)+B=2n+2x+1\iff A=1\land A+B=2x+1\iff A=1\land B=2x $$ giving (still for fixed $x$ and $n\in\Bbb Z$ variable) $$ p(x+n)=n^2+2nx+C=(n+x)^2+C-x^2 $$ as the general solution for each fixed $x$, and since $p$ is a polynomial, $C=c+x^2$ where $c$ is constant for all $x\in[0,1)$, so that finally $$p(x)=x^2+c.$$

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If $$p(x+1)-p(x)=\Delta p(x)=2x+1$$ Then $$p(x)=\sum{\Delta p(x)}=x^2+C$$

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