Proof that $(x-1)^2$ divides a $nx^{n+1} – (n+1)x^{n}+1$

Question

I want to solve:

Show that, in $\mathbb Q[x]$ (rationals), the polynomial $(x-1)^2$ divides a $zx^{z+1} – (z+1)x^{z}+1$, where $z \geq 1$ is any integer

I am really lost with this one, don't know how to aproach it

Answer 1

Method 1 (Derivatives): Let $f(x)=nx^{n+1}-(n+1)x^n+1$. Then $f(1)=n-(n+1)+1=0$, so $x-1|f(x)$.

Furthermore, $f'(x)=n(n+1)x^n-n(n+1)x^{n-1}$, so $f'(1)=n(n+1)-n(n+1)=0$. Thus, $(x-1)^2|f(x)$.

Method 2 (Coefficients): $(x-1)^2|f(x)$ is equivalent to $x^2|f(x+1)$, i.e., that the coefficient of $1$ and $x$ of $f(x+1)$ is $0$.

By the binomial theorem, $f(x+1)=n\sum_{i=0}^{n+1}\binom{n+1}ix^i-(n+1)\sum_{i=0}^n\binom{n}ix^i+1$. The constant term is $n\binom{n+1}0-(n+1)\binom n0+1=0$ and the coefficient of $x$ is $n\binom{n+1}1-(n+1)\binom n1=0$.

Method 3 (Induction): The case when $n=1$ is clear. Suppose the case when $n=k$ is true. When $n=k+1$, $$(k+1)x^{k+2}-(k+2)x^{k+1}+1=x(kx^{k+1}-(k+1)x^k+1)+(x-1)(x^{k+1}-1).$$

$x(kx^{k+1}-(k+1)x^k+1)$ is divisible by $(x-1)^2$ by the inductive hypothesis, and $(x-1)^2|(x-1)(x^{k+1}-1)$ is clear.

Answer 2

factorisation yields:

$$(x-1)\left(nx^n-(1+x+x^2...x^{n-1} )\right)$$

so we just have to prove that $$g(x)=nx^n-(1+x+x^2...x^{n-1}) $$ is divisible by $x-1$. Can you end it now?

show that $g(1)=0$

Answer 3

Put $y=x-1$. Then we have to prove that $y^2$ divides $$n(y+1)^{n+1}-(n+1)(y+1)^n+1$$ In other words, we have to prove that if you expand out that polynomial, the last two terms are $0$. But the last two terms of $(y+1)^{n+1}$ are $(n+1)y+1$, and the last two terms of $(y+1)^n$ are $ny+1$. So we get

$$n\{(n+1)y+1\}-(n+1)(ny+1)+1$$ which you can easily evaluate.

Answer 4

Step 1:

$\frac {nx^{n+1} – (n+1)x^{n}+1}{x-1}= \frac {nx^{n}(x-1) – (x^{n}-1)}{x-1}= nx^n-x^{n-1}-x^{n-2}-...-x-1$

$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$

Step 2:

Let $P(x)=nx^n-x^{n-1}-x^{n-2}-...-x-1$

$P(1)=n-n\times(-1)=0$