In this question, $k$ is a positive integer. Consider the following theorem: If $2 ^ { k } + 1$ is a prime, then $k$ is a power of $2 . \quad ( * )$ Which of the following statements, taken individually, is/are equivalent to (*)? I If $k$ is a power of 2 , then $2 ^ { k } + 1$ is prime. II $\quad 2 ^ { k } + 1$ is not prime only if $k$ is not a power of 2 . III A sufficient condition for $k$ to be a power of 2 is that $2 ^ { k } + 1$ is prime.

	Statement I is equivalent to (*)	Statement II is equivalent to (*)	Statement III is equivalent to (*)
A	Yes	Yes	Yes
B	Yes	Yes	No
C	Yes	No	Yes
D	Yes	No	No
E	No	Yes	Yes
F	No	Yes	No
G	No	No	Yes
H	No	No	No

Let $x$ be a real number. Which one of the following statements is a sufficient condition for exactly three of the other four statements?
A $x \geq 0$ B $x = 1$ C $x = 0$ or $x = 1$ D $x \geq 0$ or $x \leq 1$ E $\quad x \geq 0$ and $x \leq 1$

Assume vectors $\boldsymbol{a}_1, \boldsymbol{a}_2, \ldots, \boldsymbol{a}_m$ are linearly independent in a vector space $V$, where $m$ is an integer greater than or equal to 3. Obtain the condition that $m$ must satisfy in order for $\boldsymbol{a}_1 + \boldsymbol{a}_2,\ \boldsymbol{a}_2 + \boldsymbol{a}_3,\ \ldots,\ \boldsymbol{a}_{m-1} + \boldsymbol{a}_m$ and $\boldsymbol{a}_m + \boldsymbol{a}_1$ to be linearly independent.

4
Answer the following questions.

1. [(1)] Let $K$, $L$ be positive odd integers and $A$, $B$ be positive integers satisfying $KA = LB$. Show that if the remainder when $K$ is divided by $4$ equals the remainder when $L$ is divided by $4$, then the remainder when $A$ is divided by $4$ equals the remainder when $B$ is divided by $4$.
   
2. [(2)] Let $a$, $b$ be positive integers satisfying $a > b$. Show that there exist positive odd integers $K$, $L$ such that $KA = LB$ holds for $A = {}_{4a+1}C_{4b+1}$, $B = {}_{a}C_{b}$.
   
3. [(3)] Let $a$, $b$ be as in (2), and suppose further that $a - b$ is divisible by $2$. Show that the remainder when ${}_{4a+1}C_{4b+1}$ is divided by $4$ equals the remainder when ${}_{a}C_{b}$ is divided by $4$.
   
4. [(4)] Find the remainder when ${}_{2021}C_{37}$ is divided by $4$.

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Answer the following questions.
(1) The function $f ( x , y )$ with real variables $x , y$ is defined as follows:
$$f ( x , y ) = \left| \begin{array} { c c c } 1 & x _ { 1 } & y _ { 1 } \\ 1 & x _ { 2 } & y _ { 2 } \\ 1 & x & y \end{array} \right|$$
Show that the set of solutions of the equation $f ( x , y ) = 0$ is a line passing through two points $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$ on the $x y$ plane, where $x _ { 1 } \neq x _ { 2 }$.
(2) Find the value of the determinant $\left| \begin{array} { c c c } 1 & x _ { 1 } & x _ { 1 } ^ { 2 } \\ 1 & x _ { 2 } & x _ { 2 } ^ { 2 } \\ 1 & x _ { 3 } & x _ { 3 } ^ { 2 } \end{array} \right|$ in factored form.
(3) Show that there is a unique curve $y = a _ { 0 } + a _ { 1 } x + a _ { 2 } x ^ { 2 }$ passing through three points $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right) , \left( x _ { 3 } , y _ { 3 } \right)$ on the $x y$ plane, where $a _ { 0 } , a _ { 1 } , a _ { 2 }$ are constants and $x _ { 1 } , x _ { 2 } , x _ { 3 }$ are all distinct.
(4) The curve in (3) can be represented in the form $y = c _ { 1 } y _ { 1 } + c _ { 2 } y _ { 2 } + c _ { 3 } y _ { 3 }$, where each of $c _ { 1 } , c _ { 2 } , c _ { 3 }$ does not depend on $y _ { 1 } , y _ { 2 } , y _ { 3 }$. Find $c _ { 1 } , c _ { 2 } , c _ { 3 }$.
(5) Let us represent a curve $y = a _ { 0 } + a _ { 1 } x + a _ { 2 } x ^ { 2 } + a _ { 3 } x ^ { 3 } + a _ { 4 } x ^ { 4 }$ passing through five points $\left( x _ { 1 } , y _ { 1 } \right) , \ldots , \left( x _ { 5 } , y _ { 5 } \right)$ on the $x y$ plane in the form $y = c _ { 1 } y _ { 1 } + \cdots + c _ { 5 } y _ { 5 }$, where each of $c _ { 1 } , \ldots , c _ { 5 }$ does not depend on $y _ { 1 } , \ldots , y _ { 5 }$, and $x _ { 1 } , \ldots , x _ { 5 }$ are all distinct. Find $c _ { 1 }$.

6

An integer greater than or equal to 2 that has no positive divisors other than 1 and itself is called a prime number. Answer the following questions.

(1) Let $f(x) = x^3 + 10x^2 + 20x$. Find all integers $n$ such that $f(n)$ is a prime number.

(2) Let $a$, $b$ be integer constants, and let $g(x) = x^3 + ax^2 + bx$. Show that the number of integers $n$ such that $g(n)$ is a prime number is at most 3.
%% Page 7 1 Go to problem page
For a point $\mathrm{P}(x,\ y,\ 0)$ in the $xy$-plane different from the origin O, and point $\mathrm{A}(0,\ -1,\ 1)$,
$$\cos\angle\mathrm{AOP} = \frac{\overrightarrow{\mathrm{OA}}\cdot\overrightarrow{\mathrm{OP}}}{|\overrightarrow{\mathrm{OA}}||\overrightarrow{\mathrm{OP}}|} = \frac{-y}{\sqrt{2}\sqrt{x^2+y^2}}$$
Since $\angle\mathrm{AOP} \geq \dfrac{2}{3}\pi$ implies $\cos\angle\mathrm{AOP} \leq -\dfrac{1}{2}$, we have $\dfrac{-y}{\sqrt{2}\sqrt{x^2+y^2}} \leq -\dfrac{1}{2}$
$$2y \geq \sqrt{2}\sqrt{x^2+y^2}$$
Then, for $y \geq 0$, we get $4y^2 \geq 2(x^2+y^2)$, and from $y^2 - x^2 \geq 0$, $$(y+x)(y-x) \geq 0 \quad (y \geq 0) \cdots\cdots\textcircled{1}$$
Also, from $\overrightarrow{\mathrm{AO}} = (0,\ 1,\ -1)$, $\overrightarrow{\mathrm{AP}} = (x,\ y+1,\ -1)$,
$$\cos\angle\mathrm{OAP} = \frac{\overrightarrow{\mathrm{AO}}\cdot\overrightarrow{\mathrm{AP}}}{|\overrightarrow{\mathrm{AO}}||\overrightarrow{\mathrm{AP}}|} = \frac{y+2}{\sqrt{2}\sqrt{x^2+(y+1)^2+1}}$$
Since $\angle\mathrm{OAP} \leq \dfrac{\pi}{6}$ implies $\cos\angle\mathrm{OAP} \geq \dfrac{\sqrt{3}}{2}$, we have $\dfrac{y+2}{\sqrt{2}\sqrt{x^2+(y+1)^2+1}} \geq \dfrac{\sqrt{3}}{2}$
$$2(y+2) \geq \sqrt{6}\sqrt{x^2+(y+1)^2+1}$$
Then, for $y+2 \geq 0$, we get $4(y+2)^2 \geq 6\{x^2+(y+1)^2+1\}$, $$2(y^2+4y+4) \geq 3x^2+3(y^2+2y+1)+3$$ From $3x^2+y^2-2y-2 \leq 0$, we get $3x^2+(y-1)^2 \leq 3$, and thus
$$x^2 + \frac{(y-1)^2}{3} \leq 1 \quad (y \geq -2) \cdots\cdots\textcircled{2}$$
From \textcircled{1} and \textcircled{2}, the region in the $xy$-plane that P can occupy is the shaded region in the figure on the right. The boundary except for the origin is included.
[Figure: Shaded region in the $xy$-plane bounded by the ellipse $x^2+\frac{(y-1)^2}{3}\leq 1$ and the lines $y=x$, $y=-x$ with $y\geq 0$; key points marked at $y=1+\sqrt{3}$, $y=1$, $y=1-\sqrt{3}$, and $x=\pm 1$]
\subsection*{[Commentary]}
This is a basic problem on spatial vectors and regions. The amount of computation is relatively small.
%% Page 8 2 Go to problem page

(1) For $f(x) = \displaystyle\int_0^1 \frac{|t-x|}{1+t^2}\,dt\ (0 \leqq x \leqq 1)$, we have $f(x) = -\displaystyle\int_0^x \frac{t-x}{1+t^2}\,dt + \int_x^1 \frac{t-x}{1+t^2}\,dt$
$$f(x) = -\int_0^x \frac{t}{1+t^2}\,dt + x\int_0^x \frac{dt}{1+t^2} + \int_x^1 \frac{t}{1+t^2}\,dt - x\int_x^1 \frac{dt}{1+t^2} \quad \cdots\cdots\text{\textcircled{1}}$$
$$f'(x) = -\frac{x}{1+x^2} + \int_0^x \frac{dt}{1+t^2} + x\cdot\frac{1}{1+x^2} - \frac{x}{1+x^2} - \int_x^1 \frac{dt}{1+t^2} + x\cdot\frac{1}{1+x^2}$$
$$= \int_0^x \frac{dt}{1+t^2} - \int_x^1 \frac{dt}{1+t^2} \quad \cdots\cdots\text{\textcircled{2}}$$
From $0 < \alpha < \dfrac{\pi}{4}$, we have $0 < \tan\alpha < 1$, and $f'(\tan\alpha) = \displaystyle\int_0^{\tan\alpha} \frac{dt}{1+t^2} - \int_{\tan\alpha}^1 \frac{dt}{1+t^2}$.
Setting $t = \tan\theta\ \left(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\right)$, we get $dt = \dfrac{d\theta}{\cos^2\theta}$, so
$$f'(\tan\alpha) = \int_0^{\alpha} \frac{1}{1+\tan^2\theta}\cdot\frac{d\theta}{\cos^2\theta} - \int_{\alpha}^{\frac{\pi}{4}} \frac{1}{1+\tan^2\theta}\cdot\frac{d\theta}{\cos^2\theta}$$
$$= \int_0^{\alpha} d\theta - \int_{\alpha}^{\frac{\pi}{4}} d\theta = \alpha - \left(\frac{\pi}{4} - \alpha\right) = 2\alpha - \frac{\pi}{4}$$
Then, from $f'(\tan\alpha) = 0$, we get $2\alpha - \dfrac{\pi}{4} = 0$, so $\alpha = \dfrac{\pi}{8}$.

(2) From the half-angle formula, $$\tan^2\alpha = \tan^2\frac{\pi}{8} = \frac{1-\cos\dfrac{\pi}{4}}{1+\cos\dfrac{\pi}{4}} = \frac{1 - \dfrac{1}{\sqrt{2}}}{1 + \dfrac{1}{\sqrt{2}}} = \frac{\sqrt{2}-1}{\sqrt{2}+1} = (\sqrt{2}-1)^2$$
Since $\tan\dfrac{\pi}{8} > 0$, we have $\tan\alpha = \tan\dfrac{\pi}{8} = \sqrt{2}-1$.

(3) From \textcircled{2}, $f''(x) = \dfrac{1}{1+x^2} + \dfrac{1}{1+x^2} = \dfrac{2}{1+x^2} > 0$, so $f'(x)$ is strictly increasing on $0 \leqq x \leqq 1$.
Then, from (2), since $f'(\sqrt{2}-1) = 0$, the increase/decrease of $f(x)$ on $0 \leqq x \leqq 1$ is as shown in the table on the right.

$x$	$0$	$\cdots$	$\sqrt{2}-1$	$\cdots$	$1$
$f'(x)$		$-$	$0$	$+$
$f(x)$		$\searrow$		$\nearrow$

At this point, from \textcircled{1},
$$f(0) = \int_0^1 \frac{t}{1+t^2}\,dt = \frac{1}{2}\bigl[\log(1+t^2)\bigr]_0^1 = \frac{1}{2}\log 2$$
$$f(1) = -\int_0^1 \frac{t}{1+t^2}\,dt + \int_0^1 \frac{dt}{1+t^2} = -\frac{1}{2}\log 2 + \int_0^{\frac{\pi}{4}} d\theta = \frac{\pi}{4} - \frac{1}{2}\log 2$$
Then, since $0.69 < \log 2 < 0.7$, we get $f(0) - f(1) = \log 2 - \dfrac{\pi}{4} < 0$, so $f(0) < f(1)$.
Therefore, the maximum value is $f(1) = \dfrac{\pi}{4} - \dfrac{1}{2}\log 2$.
Also, the minimum value is $f(\sqrt{2}-1)$, and from \textcircled{1},
%% Page 9 $$f(\sqrt{2}-1) = -\int_0^{\sqrt{2}-1}\frac{t}{1+t^2}dt + (\sqrt{2}-1)\int_0^{\sqrt{2}-1}\frac{dt}{1+t^2} + \int_{\sqrt{2}-1}^{1}\frac{t}{1+t^2}dt$$ $$-(\sqrt{2}-1)\int_{\sqrt{2}-1}^{1}\frac{dt}{1+t^2}$$
Here, $\displaystyle\int_0^{\sqrt{2}-1}\frac{t}{1+t^2}dt = \frac{1}{2}\bigl[\log(1+t^2)\bigr]_0^{\sqrt{2}-1} = \frac{1}{2}\log(4-2\sqrt{2})$
$$\int_{\sqrt{2}-1}^{1}\frac{t}{1+t^2}dt = \frac{1}{2}\log 2 - \frac{1}{2}\log(4-2\sqrt{2})$$
$$\int_0^{\sqrt{2}-1}\frac{dt}{1+t^2} - \int_{\sqrt{2}-1}^{1}\frac{dt}{1+t^2} = \int_0^{\frac{\pi}{8}}d\theta - \int_{\frac{\pi}{8}}^{\frac{\pi}{4}}d\theta = \frac{\pi}{8} - \left(\frac{\pi}{4}-\frac{\pi}{8}\right) = 0$$
Summarizing, $f(\sqrt{2}-1) = -\dfrac{1}{2}\log(4-2\sqrt{2}) + \dfrac{1}{2}\log 2 - \dfrac{1}{2}\log(4-2\sqrt{2})$, and
$$f(\sqrt{2}-1) = \log\sqrt{2} - \log(4-2\sqrt{2}) = \log\frac{\sqrt{2}}{4-2\sqrt{2}} = \log\frac{1}{2\sqrt{2}-2} = \log\frac{\sqrt{2}+1}{2}$$

[Commentary]
This is a computation problem involving definite integrals using the relationship between differentiation and integration. Part (1) serves as a guide for part (3). However, the computation of the minimum value is somewhat involved.
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\boxed{3

\text{Go to Problem Page}}

1. [(1)] The points symmetric to the point $(a,\,b)$ with respect to the $x$-axis, $y$-axis, the line $y=x$, and the line $y=-x$ are $(a,\,-b)$, $(-a,\,b)$, $(b,\,a)$, and $(-b,\,-a)$, respectively.
   From this, the 8 points that point P, initially at $(2,\,1)$, can reach by the given rules are as follows.
   $$\text{A}(2,\,1),\quad \text{B}(2,\,-1),\quad \text{C}(-2,\,1),\quad \text{D}(-2,\,-1)$$ $$\text{E}(1,\,2),\quad \text{F}(1,\,-2),\quad \text{G}(-1,\,2),\quad \text{H}(-1,\,-2)$$
   [Figure: Coordinate plane showing points A through H at the described positions]
   
2. [(2)] Let the probabilities that P is at A, B, C, D, E, F, G, H after $n$ seconds be $a_n$, $b_n$, $c_n$, $d_n$, $e_n$, $f_n$, $g_n$, $h_n$, respectively. Then, $$a_{n+1} = \frac{1}{3}b_n + \frac{1}{3}c_n + \frac{1}{6}e_n + \frac{1}{6}h_n, \quad d_{n+1} = \frac{1}{3}c_n + \frac{1}{3}b_n + \frac{1}{6}h_n + \frac{1}{6}e_n$$
   From this, $a_{n+1} = d_{n+1}$, so for $n \geq 2$, $a_n = d_n$.
   Furthermore, since $a_1 = d_1 = 0$, we have $a_n = d_n$ $(n \geq 1)$.
   
3. [(3)] Similarly to (2), $a_n = d_n$, $b_n = c_n$, $e_n = h_n$, $f_n = g_n$, and $$a_n + b_n + e_n + f_n = \frac{1}{2}$$
   where $a_1 = 0$, $b_1 = \dfrac{1}{3}$, $e_1 = \dfrac{1}{6}$, $f_1 = 0$. In this case,
   $$a_{n+1} = \frac{2}{3}b_n + \frac{1}{3}e_n \cdots\cdots\cdots\textcircled{1}, \quad b_{n+1} = \frac{2}{3}a_n + \frac{1}{3}f_n \cdots\cdots\cdots\textcircled{2}$$
   $$e_{n+1} = \frac{2}{3}f_n + \frac{1}{3}a_n \cdots\cdots\cdots\textcircled{3}, \quad f_{n+1} = \frac{2}{3}e_n + \frac{1}{3}b_n \cdots\cdots\cdots\textcircled{4}$$
   From $\textcircled{1}+\textcircled{4}$, $a_{n+1} + f_{n+1} = b_n + e_n = \dfrac{1}{2} - (a_n + f_n)$, so
   $$a_{n+1} + f_{n+1} - \frac{1}{4} = -\left(a_n + f_n - \frac{1}{4}\right)$$
   Thus, $a_n + f_n - \dfrac{1}{4} = \left(a_1 + f_1 - \dfrac{1}{4}\right)(-1)^{n-1} = -\dfrac{1}{4}(-1)^{n-1}$, so
   $$a_n + f_n = \frac{1}{4}\{1-(-1)^{n-1}\} = \frac{1}{4}\{1+(-1)^n\} \cdots\cdots\cdots\textcircled{5}$$
   From $\textcircled{1}-\textcircled{4}$, $a_{n+1} - f_{n+1} = \dfrac{1}{3}(b_n - e_n)$; from $\textcircled{2}-\textcircled{3}$, $b_{n+1} - e_{n+1} = \dfrac{1}{3}(a_n - f_n)$, so
   $$a_{n+2} - f_{n+2} = \frac{1}{3}(b_{n+1} - e_{n+1}) = \frac{1}{9}(a_n - f_n) \cdots\cdots\cdots\textcircled{6}$$
   Note that $a_2 = \dfrac{2}{3}b_1 + \dfrac{1}{3}e_1 = \dfrac{2}{9} + \dfrac{1}{18} = \dfrac{5}{18}$, $f_2 = \dfrac{2}{3}e_1 + \dfrac{1}{3}b_1 = \dfrac{1}{9} + \dfrac{1}{9} = \dfrac{2}{9}$.
   From $\textcircled{6}$, $(a_{n+2} - f_{n+2}) - \dfrac{1}{3}(a_{n+1} - f_{n+1}) = -\dfrac{1}{3}\left\{(a_{n+1} - f_{n+1}) - \dfrac{1}{3}(a_n - f_n)\right\}$, so
   $$(a_{n+1} - f_{n+1}) - \frac{1}{3}(a_n - f_n) = \left\{(a_2 - f_2) - \frac{1}{3}(a_1 - f_1)\right\}\left(-\frac{1}{3}\right)^{n-1} = \frac{1}{18}\left(-\frac{1}{3}\right)^{n-1}$$
   From $\textcircled{6}$, $(a_{n+2} - f_{n+2}) + \dfrac{1}{3}(a_{n+1} - f_{n+1}) = \dfrac{1}{3}\left\{(a_{n+1} - f_{n+1}) + \dfrac{1}{3}(a_n - f_n)\right\}$, so
   

$-4-$ \copyright\ 電送数学舎 2024
%% Page 11
2024 Tokyo University (Science) First Semester Exam Solutions and Explanations
$$ (a_{n+1} - f_{n+1}) + \frac{1}{3}(a_n - f_n) = \left\{(a_2 - f_2) + \frac{1}{3}(a_1 - f_1)\right\}\left(\frac{1}{3}\right)^{n-1} = \frac{1}{18}\left(\frac{1}{3}\right)^{n-1} $$
Therefore, $\dfrac{2}{3}(a_n - f_n) = \dfrac{1}{18}\left(\dfrac{1}{3}\right)^{n-1} - \dfrac{1}{18}\left(-\dfrac{1}{3}\right)^{n-1}$ and so,
$$ a_n - f_n = \frac{1}{12}\left(\frac{1}{3}\right)^{n-1} - \frac{1}{12}\left(-\frac{1}{3}\right)^{n-1} = \frac{1}{4}\left\{\left(\frac{1}{3}\right)^n + \left(-\frac{1}{3}\right)^n\right\} \cdots\cdots\cdots\textcircled{7} $$
From $\textcircled{5} + \textcircled{7}$, $\;2a_n = \dfrac{1}{4}\left\{1 + (-1)^n + \left(\dfrac{1}{3}\right)^n + \left(-\dfrac{1}{3}\right)^n\right\}$, and thus,
$$ a_n = \frac{1}{8}\left\{1 + (-1)^n + \left(\frac{1}{3}\right)^n + \left(-\frac{1}{3}\right)^n\right\} $$
From the above, the probability $a_n$ that P is at the point $(2,\,1)$ after $n$ seconds from the start is:
$$ a_n = \frac{1}{8}\left\{1 + 1 + \left(\frac{1}{3}\right)^n + \left(\frac{1}{3}\right)^n\right\} = \frac{1}{4}\left\{1 + \left(\frac{1}{3}\right)^n\right\} \quad (n \text{ is even}) $$
$$ a_n = \frac{1}{8}\left\{1 - 1 + \left(\frac{1}{3}\right)^n - \left(\frac{1}{3}\right)^n\right\} = 0 \quad (n \text{ is odd}) $$

[Commentary]
This is a problem on probability and recurrence relations. For the recurrence relations (1) through (4) in part (3), various solution methods can be considered. Note that from equation \textcircled{5}, it is also fine to split $n$ into even and odd cases for processing.
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(1) For $f(x) = -\dfrac{\sqrt{2}}{4}x^2 + 4\sqrt{2} = -\dfrac{\sqrt{2}}{4}(x^2 - 16)$, the parabola $y = f(x)$ and the circle $C_t$ with center $\mathrm{C}(c(t),\ 0)$ and radius $r(t)$ share a common tangent at the point $\mathrm{P}(t,\ f(t))$ $(0 < t < 4)$.
Now, since $f'(x) = -\dfrac{\sqrt{2}}{2}x$, the direction vector $\vec{u}$ of the tangent to the parabola at point P can be written as $\vec{u} = \left(1,\ -\dfrac{\sqrt{2}}{2}t\right)$.
Also, $\overrightarrow{\mathrm{CP}} = \left(t - c(t),\ -\dfrac{\sqrt{2}}{4}t^2 + 4\sqrt{2}\right)$, and since the parabola and circle $C_t$ share a common tangent at P, we have $\vec{u} \cdot \overrightarrow{\mathrm{CP}} = 0$, giving $$\{t - c(t)\} + \left(-\frac{\sqrt{2}}{2}t\right)\left(-\frac{\sqrt{2}}{4}t^2 + 4\sqrt{2}\right) = 0, \quad t - c(t) + \frac{1}{4}t^3 - 4t = 0$$
From this, $c(t) = \dfrac{1}{4}t^3 - 3t$, and $$\{r(t)\}^2 = \mathrm{CP}^2 = \{t - c(t)\}^2 + \{f(t)\}^2 = \left(-\frac{1}{4}t^3 + 4t\right)^2 + \frac{1}{8}(t^2 - 16)^2$$ $$= \frac{t^2}{16}(t^2 - 16)^2 + \frac{1}{8}(t^2 - 16)^2 = \frac{1}{16}(t+4)^2(t-4)^2(t^2+2)$$
(2) From (1), circle $C_t$: $\left(x - \dfrac{1}{4}t^3 + 3t\right)^2 + y^2 = \dfrac{1}{16}(t+4)^2(t-4)^2(t^2+2)$, and since $C_t$ passes through the point $(3,\ a)$, $$\left(3 - \frac{1}{4}t^3 + 3t\right)^2 + a^2 = \frac{1}{16}(t+4)^2(t-4)^2(t^2+2)$$ $$a^2 = \frac{1}{16}(t-16)^2(t^2+2) - \frac{1}{16}(t^3 - 12t - 12)^2 \cdots\cdots\text{\textcircled{1}}$$
Here, let $g(t) = (t^2 - 16)^2(t^2 + 2) - (t^3 - 12t - 12)^2$, then $$g'(t) = 4t(t^2-16)(t^2+2) + 2t(t^2-16)^2 - 2(t^3-12t-12)(3t^2-12)$$ $$= 2t(t^2-16)\{2(t^2+2)+(t^2-16)\} - 6(t^3-12t-12)(t^2-4)$$ $$= 6t(t^2-16)(t^2-4) - 6(t^3-12t-12)(t^2-4)$$ $$= 6(t^2-4)\{(t^3-16t)-(t^3-12t-12)\} = -24(t+2)(t-2)(t-3)$$
Then, the increase/decrease of $g(t)$ for $0 < t < 4$ is as shown in the table on the right.

$t$	$0$	$\cdots$	$2$	$\cdots$	$3$	$\cdots$	$4$
$g'(t)$		$-$	$0$	$+$	$0$	$-$
$g(t)$	$368$	$\searrow$	$80$	$\nearrow$	$98$	$\searrow$	$-16$

Now, the equation \textcircled{1} in $t$ is: $$16a^2 = g(t) \cdots\cdots\text{\textcircled{2}}$$
Here, since $f(3) = -\dfrac{\sqrt{2}}{4}(9-16) = \dfrac{7}{4}\sqrt{2}$, the real number $a$ takes values $0 < a < \dfrac{7}{4}\sqrt{2}$, so $$0 < a^2 < \frac{49}{8}, \quad 0 < 16a^2 < 98 \cdots\cdots\text{\textcircled{3}}$$
Therefore, the number of real values of $t$ in the range $0 < t < 4$ satisfying \textcircled{2} is, from \textcircled{3},
$-6-$ \copyright\ 電送数学舎 2024
%% Page 13 $0 < 16a^2 < 80 \ (0 < a < \sqrt{5})$ のとき $t$ の個数は 1 個
$16a^2 = 80 \ (a = \sqrt{5})$ のとき $t$ の個数は 2 個
$80 < 16a^2 < 98 \ \left(\sqrt{5} < a < \dfrac{7}{4}\sqrt{2}\right)$ のとき $t$ の個数は 3 個

[Commentary]
This is a problem combining common tangent lines with applications to differential equations. The content is fundamental, but the calculations are quite involved.
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\boxed{5

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Consider the three points $A(1,\ 0,\ 0)$, $B(0,\ 1,\ 0)$, $C(0,\ 0,\ 1)$, and the midpoint $D\!\left(\dfrac{1}{2},\ 0,\ \dfrac{1}{2}\right)$ of segment $AC$. We consider the solid obtained by rotating $\triangle ABD$ once around the $x$-axis.
First, the equation of the plane containing $\triangle ABD$ is $x + y + z = 1$, and

• [(a)] The equation of side $AB$ is, for $0 \leq x \leq 1$: $$x + y = 1,\quad z = 0 \quad \cdots\cdots \textcircled{1}$$
  
• [(b)] The equation of side $AD$ is, for $\dfrac{1}{2} \leq x \leq 1$: $\quad x + z = 1,\quad y = 0 \quad \cdots\cdots \textcircled{2}$
  
• [(c)] The equation of side $BD$ is, for $0 \leq x \leq \dfrac{1}{2}$: $\quad x = z,\quad 2x + y = 1 \quad \cdots\cdots \textcircled{3}$

Now, when $\triangle ABD$ is cut by the plane $x = k$ ($0 \leq k \leq 1$), the cross-section is a line segment. Let $S(k)$ denote the area of the donut-shaped (annular) figure obtained by rotating this line segment around the $x$-axis.
(i) $0 \leq k \leq \dfrac{1}{2}$
The intersection of the plane $x = k$ with side $AB$ is $(k,\ 1-k,\ 0)$ from \textcircled{1}, and the intersection with side $BD$ is $(k,\ 1-2k,\ k)$ from \textcircled{3}, so the cross-section of $\triangle ABD$ is the line segment with these two points as endpoints.
Here, we further consider cases depending on whether the foot of the perpendicular dropped from the point $(k,\ 0,\ 0)$ to the line containing this segment ($x = k$, $y + z = 1 - k$) is $\left(k,\ \dfrac{1-k}{2},\ \dfrac{1-k}{2}\right)$, which lies on the segment or not.
(i-i) $\dfrac{1-k}{2} \leq 1 - 2k \left(0 \leq k \leq \dfrac{1}{3}\right)$
[Figure: cross-section diagram for case (i-i)]
In this case, the foot of the perpendicular is not contained in the segment, so the outer radius of the donut is $1 - k$, and the inner radius is $\sqrt{(1-2k)^2 + k^2} = \sqrt{5k^2 - 4k + 1}$, and the area is: $$S(k) = \pi\{(1-k)^2 - (5k^2 - 4k + 1)\}$$ $$= \pi(-4k^2 + 2k)$$
(i-ii) $\dfrac{1-k}{2} \geq 1 - 2k \left(\dfrac{1}{3} \leq k \leq \dfrac{1}{2}\right)$
[Figure: cross-section diagram for case (i-ii)]
In this case, the foot of the perpendicular is contained in the segment, so the outer radius of the donut is $1 - k$, and the inner radius is $\dfrac{1}{\sqrt{2}}(1-k)$, and the area is: $$S(k) = \pi\left\{(1-k)^2 - \frac{1}{2}(1-k)^2\right\} = \frac{\pi}{2}(1-k)^2$$
(ii) $\dfrac{1}{2} \leq k \leq 1$
The intersection of the plane $x = k$ with side $AB$ is $(k,\ 1-k,\ 0)$ from \textcircled{1}, and the intersection with side $AD$ is $(k,\ 0,\ 1-k)$ from \textcircled{2}, so the cross-section of $\triangle ABD$ is the line segment with these two points as endpoints.
%% Page 15 At this point, the outer radius of the donut shape is $1-k$, the inner radius is $\dfrac{1}{\sqrt{2}}(1-k)$, and its cross-sectional area is, $$S(k) = \pi\left\{(1-k)^2 - \frac{1}{2}(1-k)^2\right\} = \frac{\pi}{2}(1-k)^2$$
From (i)(ii), $S(k) = \pi(-4k^2 + 2k) \quad \left(0 \leqq k \leqq \dfrac{1}{3}\right)$
$$S(k) = \frac{\pi}{2}(1-k)^2 \quad \left(\frac{1}{3} \leqq k \leqq 1\right)$$
[Figure: coordinate system with $z$-axis vertical and $y$-axis horizontal, showing a triangular cross-section with labels $1-k$, $\frac{1-k}{2}$, and point at $y=1$, $z=1$]
From the above, the volume $V$ of the solid is,
$$V = \pi\int_0^{\frac{1}{3}}(-4k^2+2k)\,dk + \frac{\pi}{2}\int_{\frac{1}{3}}^{1}(1-k)^2\,dk$$
$$= \pi\left[-\frac{4}{3}k^3 + k^2\right]_0^{\frac{1}{3}} - \frac{\pi}{6}\left[(1-k)^3\right]_{\frac{1}{3}}^{1} = \frac{5}{81}\pi + \frac{4}{81}\pi = \frac{\pi}{9}$$

[Commentary]
This is a problem on the volume of a solid of revolution, a frequently appearing type. Despite having a simple diagram and relatively mild calculations, it takes a considerable amount of time nonetheless. This is a problem worth practicing.
%% Page 16 6 (Go to problem page)

(1) For $f(x) = x^3 + 10x^2 + 20x = x(x^2 + 10x + 20)$, we have $f(n) = n(n^2 + 10n + 20)$.
The integers $n$ for which $f(n)$ is prime are found by letting $p, q, r, s$ be primes:

• [(i)] $(n,\ n^2 + 10n + 20) = (1,\ p)$ where $f(n) = p$:
  $p = 1^2 + 10 \cdot 1 + 20 = 31$, so $f(1) = 31$ is prime.
  
• [(ii)] $(n,\ n^2 + 10n + 20) = (-1,\ -q)$ where $f(n) = q$:
  $q = -\{(-1)^2 + 10 \cdot (-1) + 20\} = -11$, so $f(-1) = -11$ is not prime.
  
• [(iii)] $(n,\ n^2 + 10n + 20) = (r,\ 1)$ where $f(n) = r$:
  From $r^2 + 10r + 20 = 1$, we get $r^2 + 10r + 19 = 0$, so no prime $r$ exists.
  
• [(iv)] $(n,\ n^2 + 10n + 20) = (-s,\ -1)$ where $f(n) = s$:
  From $(-s)^2 + 10(-s) + 20 = -1$, we get $s^2 - 10s + 21 = 0$, so $(s-3)(s-7) = 0$.
  Thus $s = 3,\ 7$ are both prime, and $f(-3) = 3$, $f(-7) = 7$.

From (i)--(iv), the integers $n$ for which $f(n)$ is prime are $n = 1,\ -3,\ -7$.

(2) Let $a, b$ be integer constants. For $g(x) = x^3 + ax^2 + bx = x(x^2 + ax + b)$, we have $g(n) = n(n^2 + an + b)$.
The integers $n$ for which $g(n)$ is prime are found by letting $p, q, r, s$ be primes:

• [(i)] $(n,\ n^2 + an + b) = (1,\ p)$ where $g(n) = p$:
  From $1^2 + a \cdot 1 + b = p$, we get $p = a + b + 1$ \textcircled{1}
  
• [(ii)] $(n,\ n^2 + an + b) = (-1,\ -q)$ where $g(n) = q$:
  From $(-1)^2 + a \cdot (-1) + b = -q$, we get $q = -(1 - a + b) = a - b - 1$ \textcircled{2}
  
• [(iii)] $(n,\ n^2 + an + b) = (r,\ 1)$ where $g(n) = r$:
  From $r^2 + ar + b = 1$, we get $r^2 + ar + b - 1 = 0$ \textcircled{3}
  
• [(iv)] $(n,\ n^2 + an + b) = (-s,\ -1)$ where $g(n) = s$:
  From $(-s)^2 + a(-s) + b = -1$, we get $s^2 - as + b + 1 = 0$ \textcircled{4}

From (i)--(iv), the number of integers $n$ for which $g(n)$ is prime is at most $1 + 1 + 2 + 2 = 6$.
First, consider the case where (iii) and (iv) hold simultaneously. From \textcircled{3} $-$ \textcircled{4}: $$r^2 - s^2 + a(r + s) - 2 = 0, \quad (r+s)(r - s + a) = 2 \hfill \textcircled{5}$$
Since $r + s \geq 4$, equation \textcircled{5} cannot hold, so (iii) and (iv) cannot hold simultaneously.
Therefore, the number of integers $n$ for which $g(n)$ is prime is at most 4. Below, we consider the case where (i)(ii)(iii) hold simultaneously, and the case where (i)(ii)(iv) hold simultaneously.

(a) Case where (i)(ii)(iii) hold simultaneously:
$p = a + b + 1$ \textcircled{1}, $q = a - b - 1$ \textcircled{2}
Let the two distinct prime solutions $r$ satisfying \textcircled{3} be $r = r_1,\ r_2\ (2 \leq r_1 < r_2)$. Then: $$r_1 + r_2 = -a \hfill \textcircled{3}', \quad r_1 r_2 = b - 1 \hfill \textcircled{3}''$$
$-10-$ \copyright\ 電送数学舎 2024
%% Page 17 From \textcircled{2}\textcircled{3}$'$ \textcircled{3}$''$, $$q = -n_1 - n_2 - n_1 n_2 - 1 - 1 = -n_1 n_2 - n_1 - n_2 - 2 < 0 \cdots\cdots\textcircled{6}$$ Since \textcircled{6} does not hold, there is no case where (i)(ii)(iii) hold simultaneously, and there is no case where the number of integers $n$ for which $g(n)$ is prime equals 4.
(b) The case where (i)(ii)(iv) hold simultaneously $$p = a + b + 1 \cdots\cdots\textcircled{1}, \quad q = a - b - 1 \cdots\cdots\textcircled{2}$$ Now, let $s = s_1,\ s_2\ (2 \leq s_1 < s_2)$ be distinct primes satisfying \textcircled{4}, then $$s_1 + s_2 = a \cdots\cdots\textcircled{4}', \quad s_1 s_2 = b + 1 \cdots\cdots\textcircled{4}''$$ From \textcircled{2}\textcircled{4}$'$ \textcircled{4}$''$, $$q = s_1 + s_2 - s_1 s_2 + 1 - 1 = -s_1 s_2 + s_1 + s_2 = 1 - (s_1 - 1)(s_2 - 1) < 0 \cdots\cdots\textcircled{7}$$ Since \textcircled{7} does not hold, there is no case where (i)(ii)(iv) hold simultaneously, and there is no case where the number of integers $n$ for which $g(n)$ is prime equals 4.
From (a)(b), the number of integers $n$ for which $g(n)$ is prime is at most 3.

[Commentary]
This is a proof problem using prime numbers as the subject. For the number of integers $n$ in part (2), referring to the concrete example in part (1), the argument proceeds in the order: at most 6 $\to$ at most 4 $\to$ at most 3.

An integer greater than or equal to 2 that has no positive divisors other than 1 and itself is called a prime number. Answer the following questions.

• [(1)] Let $f(x) = x^3 + 10x^2 + 20x$. Find all integers $n$ such that $f(n)$ is a prime number.
  
• [(2)] Let $a$, $b$ be integer constants, and let $g(x) = x^3 + ax^2 + bx$. Show that the number of integers $n$ such that $g(n)$ is a prime number is at most 3.

Binary operations $*$, $\oplus$, $\odot$ defined on the set of rational numbers
I. $a * b = a - b$ II. $a \oplus b = a + b + ab$ III. $a \odot b = \frac{a+b}{5}$
are defined as follows. Accordingly, which of these operations satisfy the associative property?
A) Only I
B) Only II
C) Only III
D) I and II
E) II and III

a, b are positive integers, p is a prime number and
$$a ^ { 3 } - b ^ { 3 } = p$$
Given this, which of the following is the equivalent of $a ^ { 2 } + b ^ { 2 }$ in terms of $p$?
A) $\frac { p + 1 } { 2 }$
B) $\frac { p + 3 } { 2 }$
C) $\frac { p + 2 } { 3 }$
D) $\frac { 2 p - 1 } { 2 }$
E) $\frac { 2 p + 1 } { 3 }$

The following are known about a set A.

• It consists of 6 consecutive odd natural numbers.
• The sum of the elements in the set equals 4 times the largest element.

Accordingly, what is the largest element of set $A$?
A) 21
B) 19
C) 17
D) 15
E) 13

ABCD is a square $\mathrm { AF } \perp \mathrm { FB }$ $\mathrm { DE } \perp \mathrm { AF }$ $| E F | = 4 \mathrm {~cm}$
Given that the area of triangle AFB in the figure is $30 \mathrm {~cm} ^ { 2 }$, what is the area of square ABCD in $\mathrm { cm } ^ { 2 }$?
A) 81
B) 100
C) 120
D) 136
E) 144

$$\frac { ( 10 ! ) ^ { 2 } - ( 9 ! ) ^ { 2 } } { 11 ! - 10 ! - 9 ! }$$
Which of the following is this operation equal to?
A) $8 !$
B) $9 !$
C) $10 !$
D) $8 \cdot 8 !$
E) $8 \cdot 9 !$

Ali; starting with the equality $x = y$ for non-zero, equal real numbers x and y, follows the following steps in order.
I. Let us multiply both sides of the equality by x: $$x ^ { 2 } = x \cdot y$$
II. Let us subtract $\mathrm { y } ^ { 2 }$ from both sides: $$x ^ { 2 } - y ^ { 2 } = x \cdot y - y ^ { 2 }$$
III. Let us factor both sides: $$( x + y ) ( x - y ) = y ( x - y )$$
IV. Let us divide both sides by $\mathrm { x } - \mathrm { y }$: $$x + y = y$$
V. Let us substitute y for x: $$2 y = y$$
As a result of these steps, Ali arrives at the conclusion "Every number equals twice itself."
Accordingly, in which of the numbered steps did Ali make an error?
A) I B) II C) III D) IV E) V

When all of the numbers 1, 2, 3, 4, 5, 6, and 7 are placed in 7 boxes with addition or subtraction symbols between them, with one number in each box, the result of the operation obtained is 4.
$$\square + \square + \square + \square + \square - \mathrm { A } - \mathrm { B } = 4$$
Accordingly, what is the product A · B?
A) 15
B) 24
C) 28
D) 30
E) 35

One interior angle of an $n$-sided regular polygon is calculated as $\frac{(n-2) \cdot 180^{\circ}}{n}$.
A triangular piece of paper is cut along the dashed lines as shown in the figure, 3 triangular pieces are removed, and a regular hexagon is obtained.
Given that the sum of the perimeters of the removed triangles is 36 units, what is the perimeter of the hexagon?
A) 18
B) 24
C) 30
D) 36
E) 42

A wooden piece in the shape of a square right prism with a square base has a base edge length equal to 2 times its height. When a cube with an edge length equal to the height of the wooden piece is removed from inside this wooden piece, the surface area of the resulting shape in the final state is 8 square units more than the surface area of the wooden piece in the initial state.
Accordingly, what is the volume of the wooden piece in the initial state in cubic units?
A) 32
B) 80
C) 108
D) 144
E) 256

The volume of a rectangular prism is equal to the product of its base area and height.
A closed glass container in the shape of a rectangular prism contains 360 cubic units of water. When the container is placed on a flat surface with different faces completely touching the surface, the height of the water is 2 units, 4 units, and 5 units respectively.
Accordingly, what is the volume of the container in cubic units?
A) 540
B) 720
C) 840
D) 960
E) 4080

A polynomial $P(x)$ with real coefficients and of degree four satisfies the inequality
$$P(x) \geq x$$
for every real number $x$.
$$\begin{aligned} & P(1) = 1 \\ & P(2) = 4 \\ & P(3) = 3 \end{aligned}$$
according to, $\mathbf{P(4)}$ is equal to what?