# Spot the mistakes? Proof of the Twin Prime conjecture and Goldbach's theorem

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 $n < 2x <p_x^2$

Q.E.D.

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.

Q.E.D

• 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 $m\gg p_x^2$. – Hagen von Eitzen Aug 9 '17 at 22:38
• nope, because n is equiv to 0 mod 2 and 0 mod 3 – Brad Graham Aug 9 '17 at 22:40
• Thanks for making this post. I love it. I will try and find a mistake. – I Said Roll Up n Smoke Adjoint Aug 9 '17 at 22:42
• @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 $n=2P+2\gg p_x^2$. – Hagen von Eitzen Aug 9 '17 at 22:46
• my first thought is that for 2 and 3 these non options cover all options primes could be. – 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 $P_x\#$.
• Are you sure? I thought it followed from $b_i\cdot \frac{P_x\#}{p_i}\equiv 1\pmod{p_i}$ that $b_i\cdot \frac{P_x\#}{p_i}=( \tilde b_i\cdot \frac{P_x\#}{p_i})+k\cdot p_i$ – Brad Graham Aug 9 '17 at 23:05
• 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 $0 \leq c_i \leq P_x\#$ – Alex Zorn Aug 9 '17 at 23:11