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Prove that $(x-1)^2\mid f(x)=nx^{n+1}-(n+1)x^n+1$

We can prove this easily by showing $f(1)=f'(1)=0$.
But is there a way to solve this problem without using calculus?

José Carlos Santos
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Asher2211
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    You can work modulo a polynomial (such as $\left(x-1\right)^2$) just as you can work modulo an integer. So you need to show that $nx^{n+1} \equiv \left(n+1\right)x^n - 1 \mod \left(x-1\right)^2$ for each $n$. Now the solution suggests itself: find a formula for the remainder of each $x^k$, and prove it by induction on $n$. – darij grinberg Feb 28 '21 at 17:42

4 Answers4

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Here is one way to prove it:

The statement is equivalent to $x^2|f(x+1)$. It is enough to show that $f(x+1)$'s constant term and coefficient of $x$ is $0$. You can calculate this from the binomial theorem.

Kenta S
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You can do long polynomial division and get that$$\frac{nx^{n+1}-(n+1)x^n+1}{x-1}=nx^n-x^{n-1}-x^{n-2}-\cdots-x-1$$(or, of course, you can just compute $(x-1)\times(nx^n-x^{n-1}-x^{n-2}-\cdots-x-1)$). And it is clear that $1$ is a root of $nx^n-x^{n-1}-x^{n-2}-\cdots-x-1$.

José Carlos Santos
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Define polynomials $p_n$ of degree $n-1$ by$$(x-1)^2p_n(x)=nx^{n+1}-(n+1)x^n+1$$for $n\ge1$, viz.$$p_1=1,\,p_2=2x+1,\,p_{n+2}=1+2xp_{n+1}-x^2p_n.$$

J.G.
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$f(x) =n((x-1)+1)^{n+1}-(n+1)((x-1)+1)^n+1;$

Binomial expansion:

1)The constant term:

$n-(n+1)+1=0$, i.e. all terms have (at least) a factor $(x-1)$.

2)Coefficient of $(x-1)$:

$n(n+1)-(n+1)n=0$, i.e. all remaining terms have (at least) a factor $(x-1)^2$, and we are done.

Peter Szilas
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