Exercise 1.3.17 (Twin primes)

Proof. If $p$ belongs to two pairs of twin primes, then $p-2,p,p+2$ are prime numbers.

Suppose that $p>5$, then $p\equiv \pm 1(\mathit{mod}6)$.

If $p\equiv 1(\mathit{mod}6)$, then $p+2\equiv 3(\mathit{mod}6)$, thus $3\mid p+2$, where $p+2>3$, therefore $p+2$ is not a prime.

If $p\equiv -1(\mathit{mod}6)$, then $p-2\equiv -3\equiv 3(\mathit{mod}6)$, thus $3\mid p-2$, where $p-2>3$, therefore $p-2$ is not a prime.

This contradiction shows that $p\le 5$. Since $2,3$ don’t belong to two such pairs, $5$ is the only prime belonging to two pairs of twin primes.

Now suppose that $n^2-1$ has just four positive divisors, i.e. $1,n-1,n+1,n^2-1$ are the only divisors of $n^2-1$. This shows that $n-1$ has just two divisors, so is prime, and similarly $n+1$ is prime. Thus $(n-1,n+1)$ is an orderd pair of twin primes.

Conversely, let $(p,q)$ be an ordered pair of twin primes, where $p<q$, so that $q=p+2$. Let $n=p+1=q-1$. Then $1,p=n-1,q=n+1,n^2-1$ are distinct divisors of $n^2-1$. Let $r$ be any prime divisor of $n^2-1$. Then $r\mid n-1=p$ or $r\mid n+1=q$, so $r=p$ or $r=q$. If $d$ is any divisor of $n^2-1$, then $d=p^i{q}^j$, where $i\ge 0,j\ge 0$.

If $j\ge 2$, then $q^2={(n+1)}^2\mid n^2-1$. This is impossible since ${(n+1)}^2=n^2+2n+1>n^2-1$.

If $i\ge 2$, then $p^2={(n-1)}^2\mid n^2-1$, thus $n-1\mid n+1$, so $p\mid q$: this is impossible because $p,q$ are distinct primes.

Therefore $d=p^i{q}^j$, where $0\le i\le 1$ and $0\le j\le 1$. Then $n^2-1$ has just four divisors $1,p=n-1,q=n+1$ and $\mathit{pq}=n^2-1$.

Let $A$ be the set of pairs of twin primes $\{p,q\}$ and $B$ the set of positive integers $n$ such that $n^2-1$ has exactly four divisors.

Then $\phi :A\to B,\psi :B\to A$, defined by

are well defined by the above argument, and $\psi \circ \phi =1_A,\phi \circ \psi =1_B$, so $\phi$ is a bijection (and $\psi ={\phi }^{-1}$). □