A polynomial $f ( x )$ has integer coefficients such that $f ( 0 )$ and $f ( 1 )$ are both odd numbers. Prove that $f ( x ) = 0$ has no integer solutions.

If 8 points in a plane are chosen to lie on or inside a circle of diameter 2 cm then show that the distance between some two points will be less than 1 cm.

Using the fact that $\sqrt { n }$ is an irrational number whenever $n$ is not a perfect square, show that $\sqrt { 3 } + \sqrt { 7 } + \sqrt { 21 }$ is irrational.

(a) Show that the area of a right-angled triangle with all side lengths integers is an integer divisible by 6.
(b) If all the sides and area of a triangle were rational numbers then show that the triangle is got by 'pasting' two right-angled triangles having the same property.

Show that $\frac { \ln ( 12 ) } { \ln ( 18 ) }$ is irrational.

a) We want to choose subsets $A _ { 1 } , A _ { 2 } , \ldots , A _ { k }$ of $\{ 1,2 , \ldots , n \}$ such that any two of the chosen subsets have nonempty intersection. Show that the size $k$ of any such collection of subsets is at most $2 ^ { n - 1 }$. b) For $n > 2$ show that we can always find a collection of $2 ^ { n - 1 }$ subsets $A _ { 1 } , A _ { 2 } , \ldots$ of $\{ 1,2 , \ldots , n \}$ such that any two of the $A _ { i }$ intersect, but the intersection of all $A _ { i }$ is empty.

Answer the following questions
(a) A natural number $k$ is called stable if there exist $k$ distinct natural numbers $a_{1}, \ldots, a_{k}$, each $a_{i} > 1$, such that $$\frac{1}{a_{1}} + \cdots + \frac{1}{a_{k}} = 1.$$ Show that if $k$ is stable then $k+1$ is also stable. Using this or otherwise, find all stable numbers.
(b) Let $f$ be a differentiable function defined on a subset $A$ of the real numbers. Define $$f^{*}(y) := \max_{x \in A}\{yx - f(x)\}$$ whenever the above maximum is finite.
For the function $f(x) = -\ln(x)$, determine the set of points for which $f^{*}$ is defined and find an expression for $f^{*}(y)$ involving only $y$ and constants.

We consider a real $\alpha$ such that for every prime number $p$, $p^\alpha$ is a natural number. We apply relation $$\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f(x+j) = f^{(n)}(x + y_n) \quad \text{(IV.1)}$$ to the function $f_\alpha(x) = x^\alpha$ and to the integer $n = \lfloor \alpha \rfloor + 1$ (where $\lfloor \cdot \rfloor$ denotes the floor function). We now choose $x \in \mathbb{N}^*$.
Show that the expression $$\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f_\alpha(x+j)$$ is a relative integer.

Show that $n^4 + 4^n$ is composite for all integers $n > 1$.

A natural number $n$ is said to be a perfect square when there exists a natural number $x$ satisfying $n = x ^ { 2 }$. Similarly, $n$ is said to be a perfect cube when there exists a natural number $x$ satisfying $n = x ^ { 3 }$.
In the following two cases, find the natural number $n$ that satisfies the conditions.
(i) $n$ is a perfect square. Furthermore, the number obtained by adding 13 to $n$ is also a perfect square.
(ii) $n$ is a perfect cube. Furthermore, the number obtained by adding 61 to $n$ is also a perfect cube.
First, consider (i). From the definition of a perfect square number, $n$ can be expressed as $n = x ^ { 2 }$, where $x$ is a natural number. In addition, there exists a natural number $y$ such that
$$x ^ { 2 } + 13 = y ^ { 2 } .$$
Since $x < y$, $y - x = \square \mathbf { J }$ and $y + x = \mathbf { K L }$. It follows that
$$x = \mathbf { M } , \quad y = \mathbf { N } ,$$
and finally that $n = \mathbf { O P }$.
Next, consider (ii). Similar to (i), in (ii), there exists a natural number $x$ such that $n = x ^ { 3 }$, and there also exists a natural number $y$ such that
$$x ^ { 3 } + 61 = y ^ { 3 } .$$
When we solve this equation, we obtain
$$x = \mathbf { Q } , \quad y = \mathbf { R } ,$$
and hence the perfect cube $n = \mathbf{ST}$.

Consider integers $a$ and $b$ satisfying the equation $$14a + 9b = 147. \tag{1}$$
(1) We are to find the positive integers $a$ and $b$ satisfying equation (1).
Since $$14a = \mathbf{A}(\mathbf{BC} - \mathbf{D}b) \text{ and } 9b = \mathbf{E}(\mathbf{FG} - \mathbf{H}a),$$ $a$ is a multiple of $\mathbf{A}$, and $b$ is a multiple of $\mathbf{E}$.
So, when we set $a = \mathbf{A}m$ and $b = \mathbf{E}n$, where $m$ and $n$ are integers, from (1) we have $$\mathbf{I}m + \mathbf{J}n = \mathbf{K}.$$
Since the positive integers $m$ and $n$ satisfying this are $$m = \mathbf{L} \text{ and } n = \mathbf{M},$$ we obtain $$a = \mathbf{N} \text{ and } b = \mathbf{O}.$$
(2) We are to find the solutions $a$ and $b$ of equation (1) satisfying $0 < a + b < 5$.
Since $14 \times \mathbf{N} + 9 \times \mathbf{O} = 147$, from this equality and (1) we have $$14(a - \mathbf{N}) = 9(\mathbf{O} - b).$$
Since 14 and 9 are relatively prime, $a$ and $b$ can be expressed in terms of an integer $k$ as $$a = \mathbf{P} + \mathbf{Q}k, \quad b = -\mathbf{RS}k + \mathbf{T}.$$
Since $0 < a + b < 5$, we have $k = \mathbf{U}$, and we obtain $$a = \mathbf{VW}, \quad b = -\mathbf{XY}.$$

2. For ALL APPLICANTS.

(i) Expand and simplify
$$( a - b ) \left( a ^ { n } + a ^ { n - 1 } b + a ^ { n - 2 } b ^ { 2 } + \cdots + a b ^ { n - 1 } + b ^ { n } \right)$$
(ii) The prime number 3 has the property that it is one less than a square number. Are there any other prime numbers with this property? Justify your answer.
(iii) Find all the prime numbers that are one more than a cube number. Justify your answer.
(iv) Is $3 ^ { 2015 } - 2 ^ { 2015 }$ a prime number? Explain your reasoning carefully.
(v) Is there a positive integer $k$ for which $k ^ { 3 } + 2 k ^ { 2 } + 2 k + 1$ is a cube number? Explain your reasoning carefully.
If you require additional space please use the pages at the end of the booklet

Suppose that we want to find a nice set of two numbers that are the same colour. By considering the possibilities for the colours of the numbers 1, 2, and 3, prove such a set always exists.

Suppose that we want to find a nice set of four numbers that are all the same colour. By considering the possibilities for the colours of the numbers from 1 to 27 inclusive, prove such a set always exists. [Hint: consider $\{ 1,2,3 \} , \{ 4,5,6 \} , \ldots , \{ 25,26,27 \}$.]

Consider the statement:
() A whole number $n$ is prime if it is 1 less or 5 less than a multiple of 6 .
How many counterexamples to () are there in the range $0 < n < 50$ ?

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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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.
%% Page 12 4 Go to problem page
(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

\text{Go to Problem Page}}
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.