The Twin Prime Conjecture

For any prime number $p_x$ larger than 3, there exists a number $n$ that is less than $p_x^2 -2$ and does not have a remainder of $\pm 1$ when divided by any prime number less than or equal to $p_x$.

Why is this the same as the Twin Prime Conjecture?

If $n$ exists as above, then $n \pm 1$ are prime numbers because; they are not divisible by any prime number less than or equal to $p_x$ and they are less than $p_{x+1}^2$ which is the smallest composite number whose prime factors are all greater than $p_x$. As there are infinitely many prime numbers, it follows that there are infinitely twin prime numbers.

Direct Proof

If $n$ does exist as above. It follows that $n \equiv 0\mod 2$ and $n \equiv 0 \mod 3$ as all other values contradict the OP statement. It also follows that for any other prime number $p_i$, there are only $p_i - 2$ possible remainders that $n$ can have when divided by $p_i$, because the remainders "$\pm 1$" also contradict the OP statement.

The Chinese Remainder Theorem ensures the existence of a number $m$ that satisfies the system of conditions, $m \equiv a_i\mod p_i$, where $1 \leq i \leq x$ and $a_i$ are specified remainders.

The Chinese Remainder Theorem also gives a way of calculating the value of $m$. By the Chinese Remainder Theorem, if $n$ exists as above, then

$n \equiv \Sigma_{i=3}^x (a_i)(\frac{b_iP_x\#}{p_i}) \mod P_x\#$ where $P_x\# := \Pi_{i=1}^x p_i $ and $b_i$ is given by solving $\frac{b_iP_x\#}{p_i} \equiv 1 \mod p_i$.

Let $c_i = \frac{b_iP_x\#}{p_i}$, then $n \equiv \Sigma_{i=3}^x (a_i)(c_i) \mod P_x\#$

If $c_i \equiv 1 \mod p_i$ then $c_i \mod P_x\#$ can take any one of $\frac{P_x\#}{p_i}$, depending on which value of $b_i$ is chosen.

Choose $b_i$ such that $c_i \equiv 1 \mod P_x\#$

It follows that $n \equiv \Sigma_{i=3}^x (a_i) \mod P_x\#$.

Choose $a_i = 2$, then $n < 2x <p_x^2 $


Goldbach's Conjecture

For any number $n$ such that $p_x^2 < 2n < p_{x+1}^2$ and $p_{x+1}^2 < P_x\#$, there exists another number $e < n - p_x$ such that $n \pm e$ are not divisible by any prime number less than or equal to $p_x$.

Direct Proof

For any prime number $p_i$ less than or equal to $p_x$, there are only certain remainders that $e$ can have when divided by $p_i$ to ensure that $n \pm e$ is not divisible by $p_i$.

If $e$ exists, it follows from the reasoning in the last proof that $e \equiv \Sigma_{i=1}^x (a_i) \mod P_x\#$.

For all $p_i$ that are not factors of $n$, choose $a_i = 0$. If $p_{x+1}^2 < P_x\#$ then there will be atleast one value of $i : a_i = 0$ because otherwise $p_{x+1}^2 < n$ which is a contradiction.

$\Sigma_{i=1}^x p_i < n - p_x$ for all $n \geq 14$. Starting at $n=14$, when the LHS sum increases by $p_x + \theta$, the RHS increases by an increment of $\theta p_x + \frac{1}{2}\theta^2$.

As $a_i < p_i$, and $\Sigma_{i=1}^x p_i < n - p_x$ it follows that an $e$ exists that satisfies all the conditions.


  • 1
    $\begingroup$ If you choose $a_i=2$ for all $i$, then $m\equiv 2\pmod {p_i}$ for all $i$ implies $m\equiv 2\pmod{\prod p_i}$. Then either $m=2$ or $m\gg p_x^2$. $\endgroup$ – Hagen von Eitzen Aug 9 '17 at 22:38
  • $\begingroup$ nope, because n is equiv to 0 mod 2 and 0 mod 3 $\endgroup$ – Brad Graham Aug 9 '17 at 22:40
  • 1
    $\begingroup$ Thanks for making this post. I love it. I will try and find a mistake. $\endgroup$ – Shine On You Crazy Diamond Aug 9 '17 at 22:42
  • $\begingroup$ @BradGraham OK, I forgot to remove the first primes. So your $n$ is $\equiv 0\pmod 6$ and $\equiv 2\pmod P$, where $P$ is the product over all other primes up to $p_x$. This still does not entail $n<p_x^2$. In fact, if $P\equiv 1\pmod 6$, then $n=4P+2\gg p_x^2$, and if $P\equiv -1\pmod 6$, then $n=2P+2\gg p_x^2$. $\endgroup$ – Hagen von Eitzen Aug 9 '17 at 22:46
  • 1
    $\begingroup$ my first thought is that for 2 and 3 these non options cover all options primes could be. $\endgroup$ – user451844 Aug 9 '17 at 22:55

Your essential problem is that $b_i\cdot \frac{P_x\#}{p_i}\equiv 1\pmod{p_i}$ determines $b_i$ up to a multiple of $p_i$, so $b_i=\tilde b_i+kp_i$. Hence the possible values of $c_i:=b_i\cdot \frac{P_x\#}{p_i}$ are $\tilde b_i\cdot \frac{P_x\#}{p_i}+kP_x\#$ and so contrary to your argument does not take different values modulo $P_x\#$.

  • $\begingroup$ Are you sure? I thought it followed from $b_i\cdot \frac{P_x\#}{p_i}\equiv 1\pmod{p_i}$ that $b_i\cdot \frac{P_x\#}{p_i}=( \tilde b_i\cdot \frac{P_x\#}{p_i})+k\cdot p_i$ $\endgroup$ – Brad Graham Aug 9 '17 at 23:05
  • 1
    $\begingroup$ In fact, $c_i$ is the unique solution to $c_i \equiv 1 (\text{mod } p_i)$ and $c_i \equiv 0 (\text{mod } p_j)$, $j \neq i$, with $0 \leq c_i \leq P_x\#$ $\endgroup$ – Alex Zorn Aug 9 '17 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.