1. 詹姆逊, 素数定理.
2. E. Stein and R. Shakarchi, Fourier Analysis. 导论, 普林斯顿分析讲座 I, (2003), 第7章和第8章.
Analysis I, (2003), Chapters 7, and 8.
3. E. Stein and R. Shalarchi, Complex Analysis, Princeton Lectures in Analysis II, (2003)、
第6章和第7章。
Problem 1.

Solve the equation $y^2+4=x^3$ for integers $x, y$.

Solution. We first consider the case where $y$ is even. It follows that $x$ must also be even, which implies that $x^3 \equiv 0(\bmod 8)$. Now, $y$ is congruent to 0 or $2(\bmod 4)$. If $y \equiv 0(\bmod 4)$, then $y^2+4 \equiv 4(\bmod 8)$, so we can rule out this case. However, if $y \equiv 2(\bmod 4)$, then $y^2+4 \equiv 0(\bmod 8)$. Writing $y=2 Y$ with $Y$ odd, and $x=2 X$, we have $4 Y^2+4=8 X^3$, so that
$$Y^2+1=2 X^3$$
and
$$(Y+i)(Y-i)=2 X^3=(1+i)(1-i) X^3 .$$
We note that $Y^2+1 \equiv 2(\bmod 4)$ and so $X^3$ is odd. Now,
\begin{aligned} X^3 & =\frac{(Y+i)(Y-i)}{(1+i)(1-i)} \ & =\left(\frac{1+Y}{2}+\frac{1-Y}{2} i\right)\left(\frac{1+Y}{2}-\frac{1-Y}{2} i\right) \ & =\left(\frac{1+Y}{2}\right)^2+\left(\frac{1-Y}{2}\right)^2 . \end{aligned}
We shall write this last sum as $a^2+b^2$. Since $Y$ is odd, $a$ and $b$ are integers. Notice also that $a+b=1$ so that $\operatorname{gcd}(a, b)=1$. We now have that
$$X^3=(a+b i)(a-b i)$$
We would like to establish that $(a+b i)$ and $(a-b i)$ are relatively prime. We assume there exists some nonunit $d$ such that $d \mid(a+b i)$ and $d \mid(a-b i)$. But then $d \mid[(a+b i)+(a-b i)]=2 a$ and $d \mid(a+b i)-(a-b i)=2 b i$. Since $\operatorname{gcd}(a, b)=1$, then $d \mid 2$, and thus $d$ must have even norm. But then it is impossible that $d \mid(a+b i)$ since the norm of $(a+b i)$ is $a^2+b^2=X^3$ which is odd. Thus $(a+b i)$ and $(a-b i)$ are relatively prime, and each is therefore a cube, since $\mathbb{Z}[i]$ is a unique factorization domain. We write
$$a+b i=(s+t i)^3=s^3-3 s t^2+\left(3 s^2 t-t^3\right) i .$$

Comparing real and imaginary parts yields
\begin{aligned} & a=s^3-3 s t^2, \ & b=3 s^2 t-t^3 . \end{aligned}
Adding these two equations yields $a+b=s^3-3 s t^2+3 s^2 t-t^3$. But $a+b=1$, so we have
\begin{aligned} 1 & =s^3-3 s t^2+3 s^2 t-t^3 \ & =(s-t)\left(s^2+4 s t+t^2\right) . \end{aligned}
Now, $s, t \in \mathbb{Z}$ so $(s-t)= \pm 1$ and $\left(s^2+4 s t+t^2\right)= \pm 1$. Subtracting the second equation from the square of the first we find that $-6 s t=0$ or 2 . Since $s$ and $t$ are integers, we rule out the case $-6 s t=2$ and deduce that either $s=0$ or $t=0$. Thus either $a=1, b=0$ or $a=0, b=1$. It follows that $Y= \pm 1$, so the only solutions in $\mathbb{Z}$ to the given equation with $y$ even are $x=2, y= \pm 2$.

Next, we consider the case where $y$ is odd. We write $x^3=(y+2 i)(y-2 i)$. We can deduce that $(y+2 i)$ and $(y-2 i)$ are relatively prime since if $d$ divided both, $d$ would divide both their sum and their difference, i.e., we would have $d \mid 2 y$ and $d \mid 4 i$. But then $d$ would have even norm, and since $y$ is odd, $(y+2 i)$ has odd norm; thus $d$ does not divide $(y+2 i)$. Hence, $(y+2 i)$ is a cube; we write
$$y+2 i=(q+r i)^3=q^3-3 q r^2+\left(3 q^2 r-r^3\right) i .$$
Comparing real and imaginary parts we have that $2=3 q^2 r-r^3$ so that $r \mid 2$, and the only values $r$ could thus take are \pm 1 and \pm 2 . We get that the only possible pairs $(q, r)$ we can have are $(1,1),(-1,1),(1,-2)$, and $(-1,-2)$. Solving for $y$, and excluding the cases where $y$ is even, we find that $x=5, y= \pm 11$ is the only possible solution when $y$ is odd.

Problem 2.

Show that Liouville’s theorem holds for $\alpha$ where $\alpha$ is a complex algebraic number of degree $n \geq 2$.

Solution. First we note that if $\alpha$ is algebraic, then so is $\bar{\alpha}$ (the complex conjugate of $\alpha$ ), since they satisfy the same minimal polynomial. Also, every element in an algebraic number field is algebraic, since if the field $\mathbb{Q}(\gamma)$ has degree $n$ over $\mathbb{Q}$, then for any $\beta \in \mathbb{Q}(\gamma)$ the elements $1, \beta, \ldots, \beta^n$ are surely linearly dependent. This implies that $\alpha+\bar{\alpha}=2 \operatorname{Re}(\alpha)$ and $\alpha-\bar{\alpha}=2 i \operatorname{Im}(\alpha)$ are algebraic, since they are both in the field $\mathbb{Q}(\alpha, \bar{\alpha})$.

We can apply Liouville’s theorem to $\operatorname{Re}(\alpha)$ to get a constant $c=$ $c(\operatorname{Re}(\alpha))$ such that
$$\left|\operatorname{Re}(\alpha)-\frac{p}{q}\right| \geq \frac{c}{q^m}$$
where $\operatorname{Re}(\alpha)$ has degree $m$. Now,
\begin{aligned} \left|\alpha-\frac{p}{q}\right| & =\sqrt{\left(\operatorname{Re}(\alpha)-\frac{p}{q}\right)^2+(\operatorname{Im}(\alpha))^2} \ & \geq\left|\operatorname{Re}(\alpha)-\frac{p}{q}\right| \ & \geq \frac{c}{q^m} . \end{aligned}
To prove the result, it remains only to show that if the degree of $\alpha$ is $n$, then the degree of $\operatorname{Re}(\alpha) \leq n$. Consider the polynomial
$$g(x)=\prod_{i=1}^n\left[2 x-\left(\alpha^{(i)}+\bar{\alpha}^{(i)}\right)\right],$$
where $\alpha=\alpha^{(1)}, \alpha^{(2)}, \ldots, \alpha^{(n)}$ are the algebraic conjugates of $\alpha$. Certainly $\operatorname{Re}(\alpha)$ satisfies this equation, so we must verify that its coefficients are in $\mathbb{Q}$.

To prove this, we need some Galois Theory. Let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and let $F$ be the splitting field of this polynomial (i.e., the normal closure of $\mathbb{Q}(\alpha)$ ). Recall that $f$ is also the minimal polynomial of $\bar{\alpha}$, and so $F$ contains $\alpha^{(i)}$ and $\bar{\alpha}^{(i)}$ for $i=1, \ldots, n$. Consider the Galois group of $F$, that is, all automorphisms of $F$ leaving $\mathbb{Q}$ fixed. These automorphisms permute the roots of $f$, which are simply the conjugates of $\alpha$. It is easy to see that the coefficients of $g(x)$ will remain unchanged under a permutation of the $\alpha^{(i)}$ ‘s, and so they must lie in the fixed field of the Galois group, which is $\mathbb{Q}$.

Since $\operatorname{Re}(\alpha)$ satisfies a polynomial with coefficients in $\mathbb{Q}$ of degree $n$, it follows that the minimal polynomial of $\operatorname{Re}(\alpha)$ must divide this polynomial, and so have degree less than or equal to $n$. This proves Liouville’s theorem for complex algebraic numbers.

Problem 3.

Prove that each member of the set of $n-1$ consecutive integers
$$n !+2, n !+3, \ldots, n !+n$$
is divisible by a prime which does not divide any other member of the set.

Prove that each member of the set of $n-1$ consecutive integers
$$n !+2, n !+3, \ldots, n !+n$$
is divisible by a prime which does not divide any other member of the set.
Proof. For each $2 \leq k \leq n$, consider the three cases.

• Suppose $k$ is prime and $k>n / 2$. Then $k \mid n !+k$ and $k \nmid n !+j$ for $j \neq k$ since $n !+2 k>n !+j$. In this case we are done.
• Suppose $k$ is prime and $k \leq n / 2$. Then since $2 k \leq n$, for any prime $p \leq n, p \mid n ! / k$ and hence $p \nmid n ! / k+1$. It follows that $n ! / k+1$ has a prime factor larger than $n$. This implies $n !+k$ does as well.
• Suppose $k$ is composite. Similar to the above case, for any prime $p \leq n, p \mid n ! / k$. Thus $p \nmid n ! / k+1$ and so $n !+k$ has a prime factor larger than $n$.

Now suppose $k$ is not a prime larger than $n / 2$. Let $p_k>n$ be a prime dividing $n !+k$. Since $|(n !+j)-(n !+k)|<n$ we have
$$n !+j \not \equiv n !+k \bmod p_k \quad \text { for } j \neq k .$$
Therefore $p_k \mid n !+k$ and $p_k \nmid n !+j$ for $j \neq k$.

Problem 4.

Let $\lambda_a(n)=\sum_{d \mid n} d^a \lambda(d)$ where $\lambda(n)$ is Liouville’s function. Prove that if $\sigma>$ $\max {1, \operatorname{Re}(a)+1}$, we have
$$\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\frac{\zeta(s) \zeta(2 s-2 a)}{\zeta(s-a)}$$
and
$$\sum_{n=1}^{\infty} \frac{\lambda(n) \lambda_a(n)}{n^s}=\frac{\zeta(2 s) \zeta(s-a)}{\zeta(s)} .$$

E Let $\lambda_a(n)=\sum_{d \mid n} d^a \lambda(d)$ where $\lambda(n)$ is Liouville’s function. Prove that if $\sigma>\max {1, \operatorname{Re}(a)+1}$, we have
$$\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\frac{\zeta(s) \zeta(2 s-2 a)}{\zeta(s-a)}$$
and
$$\sum_{n=1}^{\infty} \frac{\lambda(n) \lambda_a(n)}{n^s}=\frac{\zeta(2 s) \zeta(s-a)}{\zeta(s)} .$$
Proof. Since $\lambda_a=u * N^a \lambda$, for $\sigma>\max {1, \operatorname{Re}(a)+1}$, by Theorem 11.5,
$$\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\left(\sum_{n=1}^{\infty} \frac{1}{n^s}\right)\left(\sum_{n=1}^{\infty} \frac{n^a \lambda(n)}{n^s}\right)=\zeta(s) \sum_{n=1}^{\infty} \frac{\lambda(n)}{n^{s-a}} .$$
then shows
$$\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\frac{\zeta(s) \zeta(2 s-2 a)}{\zeta(s-a)} .$$
Looking at the second sum, observe $\lambda$ is completely multiplicative and $\lambda\left(d^2\right)=1$ for all $d$. Hence if $d \mid n$,
$$\lambda\left(\frac{n}{d}\right)=\lambda\left(\frac{n}{d}\right) \lambda\left(d^2\right)=\lambda(n d) .$$
From here we find
$$\lambda(n) \lambda_a(n)=\sum_{d \mid n} d^a \lambda(n d)=\sum_{d \mid n} d^a \lambda\left(\frac{n}{d}\right)=\left(N^a * \lambda\right)(n) .$$
Thus for $\sigma>\max {1, \operatorname{Re}(a)+1}$, by Theorem 11.5,
$$\sum_{n=1}^{\infty} \frac{\lambda(n) \lambda_a(n)}{n^s}=\left(\sum_{n=1}^{\infty} \frac{n^a}{n^s}\right)\left(\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s}\right)=\frac{\zeta(2 s) \zeta(s-a)}{\zeta(s)}$$