Question: Find all polynomials $p(x)$ so that $p(x+c) = p(x) + c$. My original thought was that the function forms an arithmatic sequence with all $x$ that are multiples of $c$, but I can't find a way to make my proof that the polynomial must be linear rigorous. Can someone help point out a way I can finish?
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Hint. Differentiation – Riemann Feb 17 '23 at 22:53
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1Hint: what do you get when you substitute $x=0$? – Dustan Levenstein Feb 17 '23 at 22:53
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Right. But is there any other way? I'm trying to find a method using only basic algebra. Perhaps induction? But i can't find anything. – Celwelf Feb 17 '23 at 22:58
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3Hint: Define $q(x) = p(x)-x$. Show that $q(x)$ has the same value at infinitely many $x$. The only polynomials with such a property are constant (degree $0$) polynomials. – aschepler Feb 17 '23 at 23:05
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2You already found that (for your given fixed $c$) $p(nc)=p(0)+nc$ for every integer $n\ge0.$ What does this tell you about the polynomial $p(x)-x-p(0)?$ – Anne Bauval Feb 17 '23 at 23:08
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Thank y'all i really appreciate it. – Celwelf Feb 19 '23 at 03:01
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Assume you have two polynomial solutions $p$ and $q$ and call $f=p-q$ then it verifies $f(x+c)=f(x)$.
$f$ is a polynomial, so it cannot be periodic unless constant, there are many ways to prove this, for instance polynomials of degree $\ge 1$ are unbounded while continuous periodic functions are bounded.
As a consequence $\exists a\in\mathbb R\mid f(x)=a$ constant and $p(x)=q(x)+a$
Since $p(x)=x$ is a trivial solution, then all solutions are of the form $x+a$.
zwim
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