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# 数学代写|代数数论代写Algebraic Number Theory代考|MATH160 The ABC Conjecture

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## 数学代写|代数数论代写Algebraic Number Theory代考|The ABC Conjecture

Exercise 1.3.1 Assuming the $A B C$ Conjecture, show that if $x y z \neq 0$ and $x^n+$ $y^n=z^n$ for three mutually coprime integers $x, y$, and $z$, then $n$ is bounded.
Solution. First observe that $\max (|x|,|y|,|z|)>1$ for otherwise we have $x y z=0$. By the $A B C$ Conjecture, we have
$$\max \left(|x|^n,|y|^n,|z|^n\right) \leq \kappa(\varepsilon)(\operatorname{rad}(x y z))^{1+\varepsilon} .$$
Without any loss of generality, suppose that $\max (|x|,|y|,|z|)=|z|$. We deduce that $|z|^n \leq \kappa(\varepsilon)|z|^{3+3 \varepsilon}$. Since $|z|>1$, we conclude that $n$ is bounded.
Exercise 1.3.2 Let $p$ be an odd prime. Suppose that $2^n \equiv 1(\bmod p)$ and $2^n \not \equiv 1\left(\bmod p^2\right)$. Show that $2^d \not \equiv 1\left(\bmod p^2\right)$ where $d$ is the order of $2(\bmod p)$.

Solution. Since $2^n \equiv 1(\bmod p)$, we must have $d \mid n$. Write $n=$ de. If $2^d=1+k p$ and $p \mid k$, then
\begin{aligned} 2^n=2^{\text {de }} & =(1+k p)^e \ & \equiv 1+k p e \quad\left(\bmod p^2\right) \ & \equiv 1 \quad\left(\bmod p^2\right) \end{aligned}
a contradiction. This proves the result.

## 数学代写|代数数论代写Algebraic Number Theory代考|Supplementary Problems

Exercise 1.4.1 Show that every proper ideal of $\mathbb{Z}$ is of the form $n \mathbb{Z}$ for some integer $n$.

Solution. Suppose there is an ideal $I$ for which this is not true. Then show that there exist elements $a, b \in I$ such that $\operatorname{gcd}(a, b)=1$.

Exercise 1.4.2 An ideal $I$ is called prime if $a b \in I$ implies $a \in I$ or $b \in I$. Prove that every prime ideal of $\mathbb{Z}$ is of the form $p \mathbb{Z}$ for some prime integer $p$.

Solution. If $I$ is an ideal, then it is of the form $n \mathbb{Z}$ for some integer $n$ by the previous question. Then $a b \in I$ implies that $n \mid a b$. But then since $I$ is prime, either $a \in I$ or $b \in I$, so $n \mid a$ or $n \mid b$, implying that $n$ is prime.
Exercise 1.4.3 Prove that if the number of prime Fermat numbers is finite, then the number of primes of the form $2^n+1$ is finite.

Solution. Consider primes of the form $2^n+1$. If $n$ has an odd factor, say $n=r s$ with $r$ odd, then $2^{r s}+1$ is divisible by $2^s+1$, and is therefore not prime.

Exercise 1.4.4 If $n>1$ and $a^n-1$ is prime, prove that $a=2$ and $n$ is prime.
Solution. If $a>2$, then $a^n-1$ is divisible by $a-1$. So assume $a=2$. Then if $n$ has a factor, say $k$, then $2^k-1 \mid 2^n-1$. Therefore if $a^n-1$ is prime, $a=2$ and $n$ is prime. Numbers of this form are called Mersenne numbers.

## 数学代写|代数数论代写Algebraic Number Theory代考|The ABC Conjecture

$$\max \left(|x|^n,|y|^n,|z|^n\right) \leq \kappa(\varepsilon)(\operatorname{rad}(x y z))^{1+\varepsilon} .$$

$$2^n=2^{\mathrm{de}}=(1+k p)^e \quad \equiv 1+k p e \quad\left(\bmod p^2\right) \equiv 1 \quad\left(\bmod p^2\right)$$

## MATLAB代写

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