\noindent\textbf{6} \hfill (See solution page)

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\noindent Let $b, c, p, q, r$ be constants such that $x^4 + bx + c = (x^2 + px + q)(x^2 - px + r)$ is an identity in $x$.

\medskip
\noindent (1) When $p \neq 0$, express $q, r$ in terms of $p, b$.

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\noindent (2) Let $p \neq 0$. When $b, c$ are expressed using a constant $a$ as $b = (a^2+1)(a+2)$, $c = -\left(a + \dfrac{3}{4}\right)(a^2+1)$, find one pair of polynomials $f(t)$ and $g(t)$ in $t$ with rational coefficients satisfying
$$\{p^2 - (a^2+1)\}\{p^4 + f(a)\,p^2 + g(a)\} = 0.$$

\medskip
\noindent (3) Let $a$ be an integer. Find all integers $a$ such that the degree-4 polynomial in $x$
$$x^4 + (a^2+1)(a+2)x - \left(a+\frac{3}{4}\right)(a^2+1)$$
can be factored into a product of two quadratic polynomials with rational coefficients.

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$-6-$
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\noindent\textbf{1} \hfill \textit{Go to problem page}

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\noindent(1) Solving the parabola $C: y = x^2 + ax + b \cdots\textcircled{1}$ and the parabola $y = -x^2 \cdots\textcircled{2}$ simultaneously,
$$x^2 + ax + b = -x^2, \quad 2x^2 + ax + b = 0 \cdots\cdots\textcircled{3}$$

Since parabolas \textcircled{1}\textcircled{2} have 2 intersection points with $-1 < x < 0$ and $0 < x < 1$, the real roots of \textcircled{3} must exist in $-1 < x < 0$ and $0 < x < 1$. Setting $f(x) = 2x^2 + ax + b$,
$$f(-1) = 2 - a + b > 0, \quad f(0) = b < 0, \quad f(1) = 2 + a + b > 0$$

Summarizing: $b > a - 2$, $b > -a - 2$, $b < 0 \cdots\cdots\textcircled{4}$

From this, the range of possible values of the point $(a, b)$ is the shaded region in the figure on the right. However, the boundary lines are not included in the region.

\textit{[Figure: shaded triangular region in the $ab$-plane with vertices near $(\pm 2, 0)$ and $(0,-2)$]}

\medskip

\noindent(2) Under $(a, b)$ satisfying the system of inequalities \textcircled{4}, from \textcircled{1},
$$b = -xa - x^2 + y \cdots\cdots\textcircled{5}$$

Then, the condition for the parabola $C$ to pass through $(x, y)$ is that line \textcircled{5} has an intersection with the shaded region from (1). First, noting the $b$-intercept $-x^2 + y$ of line \textcircled{5}, when $(a, b) = (2, 0)$ we get $-x^2 + y = 2x$, when $(a, b) = (-2, 0)$ we get $-x^2 + y = -2x$, and when $(a, b) = (0, -2)$ we get $-x^2 + y = -2$.

From this, classifying by the value of the slope $-x$ of line \textcircled{5}:

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\noindent(i) $-x < -1$ $(x > 1)$:
$$-2x < -x^2 + y < 2x, \quad x^2 - 2x < y < x^2 + 2x$$

\noindent(ii) $-1 \leq -x < 0$ $(0 < x \leq 1)$:
$$-2 < -x^2 + y < 2x, \quad x^2 - 2 < y < x^2 + 2x$$

\noindent(iii) $0 \leq -x < 1$ $(-1 < x \leq 0)$:
$$-2 < -x^2 + y < -2x, \quad x^2 - 2 < y < x^2 - 2x$$

\noindent(iv) $-x \geq 1$ $(x \leq -1)$:
$$2x < -x^2 + y < -2x, \quad x^2 + 2x < y < x^2 - 2x$$

\textit{[Figures: shaded regions in $ab$-plane for cases (i),(ii),(iii),(iv)]}

\medskip

\noindent From (i)$\sim$(iv), the boundary curves of the region are:
$$y = x^2 - 2x \cdots\textcircled{6}$$
$$y = x^2 + 2x \cdots\textcircled{7}, \quad y = x^2 - 2 \cdots\textcircled{8}$$

The intersection of \textcircled{6}\textcircled{7} is $(0,\, 0)$, the intersection of \textcircled{6}\textcircled{8} is $(1,\, -1)$, and the intersection of \textcircled{7}\textcircled{8} is $(-1,\, -1)$. Therefore, the range through which parabola $C$ can pass is the shaded region in the figure on the right. However, the boundary is not included in the region.

\textit{[Figure: shaded region in the $xy$-plane bounded by the three parabolas, with key points at $(0,0)$, $(1,-1)$, $(-1,-1)$]}

\bigskip

\noindent\textbf{[Commentary]}

This is a standard problem on the region swept by a parabola. Since the conditions are given as a system of inequalities, we used a graphical approach. This is a problem worth practicing.



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\section*{\boxed{2} \hfill \text{Go to Problem Page}}

\textbf{(1)} For $f(z) = az^2 + bz + c$, from $f(0) = \alpha$, $f(1) = \beta$, $f(i) = \gamma$:
$$c = \alpha \cdots\cdots\textcircled{1}, \quad a + b + c = \beta \cdots\cdots\textcircled{2}, \quad -a + bi + c = \gamma \cdots\cdots\textcircled{3}$$

From \textcircled{1}\textcircled{2}: $a + b = \beta - \alpha$, \quad from \textcircled{1}\textcircled{3}: $-a + bi = \gamma - \alpha$, and thus:
$$b = \frac{-2\alpha + \beta + \gamma}{1+i} = \frac{(-2\alpha + \beta + \gamma)(1-i)}{2} = (-1+i)\alpha + \frac{1-i}{2}\beta + \frac{1-i}{2}\gamma$$
$$a = \beta - \alpha - (-1+i)\alpha - \frac{1-i}{2}\beta - \frac{1-i}{2}\gamma = -i\alpha + \frac{1+i}{2}\beta - \frac{1-i}{2}\gamma$$

\textbf{(2)} Under $1 \leq \alpha \leq 2$, $1 \leq \beta \leq 2$, $1 \leq \gamma \leq 2$, applying the result of (1) to $f(2) = 4a + 2b + c$:
\begin{align*}
f(2) &= -4i\alpha + 2(1+i)\beta - 2(1-i)\gamma + 2(-1+i)\alpha + (1-i)\beta + (1-i)\gamma + \alpha \\
&= (-1-2i)\alpha + (3+i)\beta + (-1+i)\gamma = (-\alpha + 3\beta - \gamma) + (-2\alpha + \beta + \gamma)i
\end{align*}

Now, setting $f(2) = x + yi$: $x = -\alpha + 3\beta - \gamma$, $y = -2\alpha + \beta + \gamma$, so:
$$(x,\ y) = \alpha(-1,\ -2) + \beta(3,\ 1) + \gamma(-1,\ 1)$$

Here, let $\vec{x} = (x,\ y)$, $\vec{a} = (-1,\ -2)$, $\vec{b} = (3,\ 1)$, $\vec{c} = (-1,\ 1)$, then:
$$\vec{x} = \alpha\vec{a} + \beta\vec{b} + \gamma\vec{c}$$

First, set $\vec{x'} = \alpha\vec{a} + \beta\vec{b}$, and let $\alpha$ and $\beta$ vary independently with $1 \leq \alpha \leq 2$, $1 \leq \beta \leq 2$. Then the range of $\vec{x'}$ is the interior or boundary of the parallelogram with the shaded lattice points shown in the figure on the right.

\textit{[Figure: parallelogram formed by vectors $\vec{a}$, $\vec{b}$ with vertices at $\vec{a}$, $2\vec{a}$, $\vec{b}$, $2\vec{b}$ combinations]}

Next, letting $\gamma$ vary independently of $\alpha$, $\beta$ with $1 \leq \gamma \leq 2$, from $\vec{x} = \vec{x'} + \gamma\vec{c}$, the range of $\vec{x}$ can be expressed as the region swept from the interior or boundary of the parallelogram of $\vec{x'}$ translated by $\vec{c}$ through to translated by $2\vec{c}$. Illustrating this gives the shaded region in the figure on the right.

\textit{[Figure: region swept by translating the parallelogram by $\vec{c}$ to $2\vec{c}$]}

From the above, the range of $f(2)$ illustrated on the complex plane is the shaded region in the figure on the right. The boundary is included in the region.

\textit{[Figure: final shaded hexagonal region on the complex plane with vertices approximately at $(-1,-2)$, $(3,1)$, $(4,-1)$, boundary included]}

\section*{[Commentary]}

This is a problem about regions on the complex plane, and is a frequently appearing type where one fixes one variable and considers the rest.

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$-2-$ \hfill \copyright\ 電送数学舎 2021
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