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Find all polynomials $p $ satisfying $p(x+1)=p(x)+2x+1. $

I found this on a local question paper, and I am unable to solve it.

Any help will be appreciated.

user333900
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  • By an easy induction $p(n)=n^2+k$, were $n\in\mathbb{N}$ and $k$ is a constant. Suppose $p(x)$ was a polynomial of degree $n$. Then it would match $x^2+k$ on more than $n+1$ values. – almagest May 02 '16 at 14:51

1 Answers1

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Given $p(x+1)-p(x) = 2x+1,$ Means $p(x) = ax^2+bx+c$

So $$a(x+1)^2+b(x+1)+c-\left[ax^2+bx+c\right] = 2x+1$$

So $2ax+a+b = 2x+1,$ Means $a=1$ and $b=0$

So we get $p(x) = x^2+c$

juantheron
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  • Omits the core of the proof, which is why the degree of any solution must be at most 2. – Did Aug 17 '16 at 08:38