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Let $a,b,c,d$ be four integers (not necessarily distinct) in the set $\{1,2,3,4,5\}$. Find the number of polynomials of the form $x^4+ ax^3 + bx^2 + cx +d$ which is divisible by $x+1$.

My Try:

Let $f(x) = x^4+ ax^3 + bx^2 + cx +d$, then $f(-1) = 0$. Thus $1+ (b+d) = c+a$. On counting cases I got 80 permissible cases. Is there a way to solve the above equation $1+ (b+d) = c+a$?

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  • $\begingroup$ You mean $1+b+d=c+\color{blue}{a}$, surely? $\endgroup$
    – J.G.
    Commented Oct 19, 2019 at 16:16

2 Answers 2

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The counting can be a bit simplified using your intermediate result as follows:

  • You have $a-b+c-d = 1 \Leftrightarrow (a-1) + (5-b) + (c-1) + (5-d) = 9$

So, your question is equivalent to counting the number of integer solutions of $$a' + b' + c' + d' = 9 \mbox{ with } a',b',c',d' \in \{0,1,2,3,4\}$$

Now, let the following sink in first by considering the exponents: This number is the same as the coefficient of $x^9$ in $(1+x+x^2+x^3+x^4)^4$.

Hence, using

  • $1+x+x^2+x^3+x^4 = \frac{1-x^5}{1-x}$ and
  • $\frac{1}{(1-x)^4} = \sum_{n=0}\binom{n+3}{3}x^n$ you get

\begin{eqnarray*}[x^9]\left(1+x+x^2+x^3+x^4\right)^4 & = & [x^9]\left(\frac{1-x^5}{1-x}\right)^4\\ & = & [x^9](1-4x^5)\sum_{n=0}\binom{n+3}{3}x^n\\ & = & \binom{9+3}{3} - 4\cdot \binom{4+3}{3}\\ & = & 80 \end{eqnarray*}

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    $\begingroup$ How and why did you think we should write a as a-1 and -b as 5-b ....? $\endgroup$
    – Zenix
    Commented Oct 19, 2019 at 17:45
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    $\begingroup$ @Zenix Because, that way the variables become all non-negative and have the same range. So, the polynomial which gives the searched for number becomes relatively simple - as you can see. $\endgroup$ Commented Oct 19, 2019 at 17:48
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The number of solutions of $a-b+c-d=1$ for $a,b,c,d\in\{1,2,3,4,5\}$ can be counted as the coefficient of $x$ in the Laurent series

$$ \left(x+x^2+x^3+x^4+x^5\right)^2\left(\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}\right)^2,$$ which is also the coefficient of $x$ in $$ \left(\frac{1}{x^4}+\frac{2}{x^3}+\frac{3}{x^2}+\frac{4}{x}+5+4x+3x^2+2x^3+x^4\right)^2, $$ so it is given by $$ 2\cdot 1+3\cdot 2+4\cdot 3+5\cdot 4+4\cdot 5+3\cdot 4+2\cdot 3+1\cdot 2=2\sum_{k=1}^{4}k(k+1)=80 $$ as claimed.

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