Problem on theory of equations.

Question

Let $P(x) := \prod_{k = 1}^{50} (x - k)$ and $Q(x) := \prod_{k = 1}^{50} (x + k)$. If $P(x) Q(x) = \sum_{k = 0}^{100} a_k x^k$, find $a_{100} - a_{99} - a_{98} - a_{97}$.

The correct answer is 42926. Can you explain the solution to this problem?

I have used the formula for the sum of roots for $P(x)Q(x)$ (Vieta's formula), and concluded $a_{99}=0$. $$ \sum_{k = 1}^{n} r_k = - \frac{a_{n - 1}}{a_n} $$

Answer 1

Let $$Q(x)=\sum_{i=1}^{50} b_ix^i.$$ Then $P(x)=Q(-x)$, so $$P(x)=\sum_{i=1}^{50} (-1)^ib_ix^i.$$

$$a_{100}=b_{50}^2=1$$ $$a_{99}=b_{50}b_{49}-b_{49}b_{50}=0$$ $$a_{98}=2b_{48}b_{50}-b_{49}^2$$ $$=\sum_{i\ne j} ij - \left(\sum_i i\right)^2=-\sum_{i=1}^{50} i^2=-42925.$$ $$a_{97}=b_{50}b_{47}-b_{49}b_{48}+b_{48}b_{49}-b_{47}b_{50}=0.$$

Thus we get $1-(-42925)=42926$ as an answer.

Answer 2

We know that the roots of $P$ are $\{1, \ldots, 50\}$ and the roots of $Q$ are $\{-1, \ldots, -50\}$. Therefore the roots of $P \cdot Q$ are $\{-50, \ldots, -2, -1, 1, 2, \ldots, 50\}$, which we'll call $r_1, \ldots, r_{100}$. From the first of Vietas formulas we know that $$ 0 = -50 + \ldots + -2 + -1 + 1 + 2 + \ldots 50 = \sum_{k = 0}^{100} r_{k} = - \frac{a_{99}}{a_{100}}. $$ This shows $a_{99} = 0$.

From the second formula we see that \begin{align} -42925 & = \sum_{k = 1}^{50} (- k^2) -50(-49 + 48 + \ldots + -1 + 1 + \ldots + 49 + 50) + (-49)(50 + 49) + \ldots + (49)(50) \\ & = -50^2 + (-49)(49 + 50) + (-48)(48 + 49 + 50) + \ldots + (49)(50)\\ & = -50^2 - 49^2 - 48^2 - \ldots - 1 \\ & = \sum_{k = 1}^{100} \sum_{\ell = k}^{100} r_k r_{\ell} = \frac{a_{98}}{a_{100}}, \end{align}

From the third formula we get that $$ \prod_{k = 1}^{50} k^2 = (-50)(50) \ldots (-2)(2)(-1)(1) = \prod_{k = 1}^{100} r_k = (-1)^{100} \frac{a_0}{a_{100}} = \frac{a_0}{a_{100}}, $$ Lastly we have $a_0 = \prod_{k = 1}^{100} r_k$ by easy inspection. Then we have $a_{100} = \frac{a_{0}}{\prod_{k = 1}^{100} r_k} = 1$ from the third formula and therefore $a_{98} = \sum_{k = 1}^{5} (-k^2) = -42925$ yielding $$ a_{100} - a_{99} - a_{98} - a_{97} = 1 - 0 - (-42925) - a_{97} = 42926 - a_{97}. $$ Now since $P(x)Q(x) = \prod_{k = 1}^{50} (x^k - k^2)$ only the even coefficients are non-zero. This proves the claim.

Answer 3

My approach would be to do a quick inspection: $$ P(x)Q(x) =\prod_{i=1}^{50} (x+i)(x-i) = \prod_{i=1}^{50} (x^2-i^2). $$ This immediate tells us two things:

1) We only need to worry about even coefficients

2) Everything will be terms of squares

Consider $a_{100}$ it must be $1$ since the only way to get to $x^{100}$ from this requires multiplying the $x$ terms together.

For $a_{98}$ we pick one term out of the product to be the non-x term and the rest are the $x$ to ensure that we get a power of 97. This gives $\sum_i i^2 = \frac{n(n+1)(2n+1)}{6} = 42925$ so $1-(-42925) = 42926$