Welcome guest
You're not logged in.
319 users online, thereof 0 logged in

## Proof: (related to "Sieve for Twin Primes")

By hypothesis, $n\ge 1$ is an integer and let $f_s:=\left\lfloor\frac {s+1}6\right\rfloor$ using the floor function.

### $”\Rightarrow”$

• Let the integer $a_n:=36n^2-1$ be the product of two twin primes $p=6n-1$ and $q=6n+1.$
• According to the fundamental theorem of arithmetics, since $36n^2-1=(6n-1)(6n+1),$ we have $p=6n-1$ and $q=6n+1.$
• Assume, there exists a prime number $s$ with $5\le s\le\sqrt q$ and $n$ congruent to $n\equiv +f_s\mod s$ or $n\equiv -f_s\mod s.$
• By assumption, either $s\mid n-f_s$ or $s\mid n+f_s.$
• Therefore, by defintion of divisibility there exist integers $a,b$ such that either $as=n-f_s$ or $bs=n+f_s.$
• Therefore, either $6as=6n-6f_s$ or $6bs=6n+6f_s.$
• Therefore, either $6as+(6f_s-1)=p$ or $6bs-(6f_s-1)=q.$
• Note1 that by definition of $f_s$, $s=6f_s-1.$
• Therefore, either $s(6a+1)=p$ or $s(6b-1)=q.$
• Therefore, either $s\mid p$ or $s\mid q.$
• Since $p,q$ are prime, we have either $s=p$ or $s=q.$
• This contradicts $s\le\sqrt{p+2}=\sqrt{q}.$
• Therefore, the assumption must be wrong and we have $n\not\equiv \pm f_s\mod s$ for all primes $s$ with $5\le s\le\sqrt{q}.$

### $”\Leftarrow”$

• Let $n\not\equiv \pm f_s\mod s$ for all primes $s$ with $5\le s\le\sqrt{q}.$
• Therefore, analogosly to $”\Rightarrow”$, there do not exist any integers $a,b$ satisfying either $s(6a+1)=6n-1=:p$ or $s(6b-1)=6n+1=:q.$
• Therefore, $s\not\mid p$ and $s\not\mid q$ for all primes $s$ with $5\le s \le\sqrt q.$
• Since, by construction of $p,q$, $pq=36n^2-1$, even $s\not\mid p$ and $s\not\mid q$ for all primes $s$ with $2\le s \le\sqrt q,$ (otherwise we would have $2\mid 36n^2-1$ or $3\mid 36n^2-1,$ which is not the case).
• Therefore, $p,q$ are prime numbers.
• Since $p=q-2,$ they are twin primes.

1 $s$ has either the form $6\nu-1$ or $6\nu+1$ for some $\nu\ge 1.$ If $s=6\nu-1$ then $6f_s-1=6\left\lfloor\frac{6\nu-1+1}{6}\right\rfloor-1=6\nu-1=s$. If $s=6\nu+1$ then $6f_s-1=6\left\lfloor\frac{6\nu+1+1}{6}\right\rfloor-1=6\lfloor\nu+\frac 13\rfloor-1=6\nu-1=s.$ In both cases, $s=6f_s-1.$

q.e.d

| | | | created: 2020-03-14 10:29:29 | modified: 2020-03-14 14:17:25 | by: bookofproofs | references: [6913]