质数问题

(1)（欧几里得）假设质数只有有限多个 $p_1,\ldots,p_n$。令

$$N=p_1p_2\cdots p_n+1.$$

则 $N$ 除以任意 $p_i$ 都余 1，因此不被任何 $p_i$ 整除。于是 $N$ 要么本身是质数，要么有某个不在列表中的质因子，矛盾。故质数无穷多。

(2) 不可能。

除 2 以外质数都为奇数，因此两个相邻质数（都大于 2）为奇数，均值

$$\frac{p+q}{2}$$

是介于 $p$ 与 $q$ 之间的整数。若它是质数，则它是位于 $p$ 与 $q$ 之间的质数，与“相邻质数”矛盾。

(3) 对 $p\ge 5$ 的质数，$p$ 为奇数且不被 3 整除。

• 模 8：任意奇数 $p\equiv 1,3,5,7\pmod 8$，平方都满足 $p^2\equiv 1\pmod 8$，所以 $8\mid(p^2-1)$。
• 模 3：因为 $p\not\equiv 0\pmod 3$，所以 $p\equiv \pm 1\pmod 3$，从而 $p^2\equiv 1\pmod 3$，所以 $3\mid(p^2-1)$。

又 $\mathrm{lcm}(8,3)=24$，因此 $\boxed{24\mid(p^2-1)}$。

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英文解析

(1) (Euclid) Assume there are only finitely many primes $p_1, \ldots, p_n$. Let
$$N=p_1p_2\cdots p_n+1.$$
Then $N$ leaves a remainder of 1 when divided by any $p_i$, so it is not divisible by any $p_i$. Thus, $N$is either itself prime or has a prime factor not in the list, a contradiction. Therefore, there are infinitely many primes.

(2) Impossible.

Except for 2, all primes are odd. Therefore, two adjacent primes (both greater than 2) are odd, and their mean
$$\frac{p+q}{2}$$
is an integer between $p$ and $q$. If it is prime, then it is a prime located between $p$and $q$, contradicting that they are "adjacent primes."

(3) For primes $p \ge 5$, $p$is odd and not divisible by 3.

• Modulo 8: For any odd prime $p \equiv 1,3,5,7 \pmod 8$, the square satisfies $p^2 \equiv 1 \pmod 8$, so $8 \mid (p^2-1)$.
• Modulo 3: Since $p \not\equiv 0 \pmod 3$, we have $p \equiv \pm 1 \pmod 3$, thus $p^2 \equiv 1 \pmod 3$, so $3 \mid (p^2-1)$.

Since $\mathrm{lcm}(8,3)=24$, we have $\boxed{24\mid(p^2-1)}$.