Find $a+b$ for $a, b$ such that $(x+1)^{n}(x^{2}+ax+b) \equiv 2^{n}(x-1) \mod (x-1)^{2}$

Question

Since $2^{n} = \sum_{0}^{n}\binom{n}{k},$ we have from the given congruence the congruence $$\sum_{0}^{n}\binom{n}{k}(x^{k+2} + ax^{k+1} + bx^{k} - x +1) \equiv 0 \mod (x-1)^{2}.$$

The given answer is $a+b = -1,$ and I noticed that for $k=0$ we "have" $$x^{2} + (a-1)x + b + 1 \equiv 0 \mod (x-1)^{2} ,$$ and by equating the coefficients we get $a=-1$ and $b=0$ so that $a+b = -1.$

However,there seems no ground for one to draw such a conclusion? Because the sum of polynomials divisible by something does not imply each individual polynomial is divisible by the thing. My question is if there is any hidden information that can be utilized to draw the above conclusion? Or simply the intended approach is misled?

A preferred solution is a solution by high-school methods.

Answer 1

We consider $P(x)=(x+1)^{n}(x^{2}+ax+b)-2^{n}(x-1)$. Then you question becomes: Find $a$ and $b$ such that $(x-1)^2$ divides $P$. But $(x-1)^2$ divides $P$ if and only if $P(1)=P'(1)=0$.

Now, $P(1)=2^n(1+a+b)$ and $P'(x)=n(x+1)^{n-1}(x^2+ax+b)+(x+1)^n(2x+a)-2^n$, so $P'(1)=n2^{n-1}(1+a+b)+2^n(1+a)$

so the problem translates to solve the system $$ \left\{ \begin{array}{rcl}1+a+b & = & 0 \\ n2^{n-1}(1+a+b)+2^n(1+a) & = & 0 \\\end{array}\right. $$

Clearly, this means that $a=-1$ and $b=0$.

Note: I am assuming that $n\geq 1$ in my answer. Case $n=1$ is easy.

Answer 2

$(x+1)^{n}(x^2+ax+b) \equiv 2^n(x-1) \mod (x-1)^2$

We must have $(x-1) \mid (x^2+ax+b)$. Hence $x^2 + ax + b = (x-1)(x+b)$

Then $(x+1)^{n}(x+b) \equiv 2^n \mod x-1$

Letting $x = 1$, we get $2^n(b+1) - 2^n = 0$

So $b = 0$ and $a = -1$