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I was re-watching some Numberphile videos on the Goldbach conjecture and other things, and in on of them it mentioned something about primes being used in Unique Factorisation. In particular, what would happen if instead of any number being the product of primes, it was instead the sum of primes.

This got me thinking about the Goldbach conjecture.

So, first I looked at the opening statement.

$$2k = P_1 + P_2$$Where $P_x \in \Bbb P$ (I shall use $\Bbb P$ as the set of prime numbers.)

The weak conjecture has already been proven, showing that any odd number can be represented as the sum of three primes.

Therefore: $$P_1 = P_3+P_4+P_5$$ $$P_2 = P_6+P_7+P_8$$ $$2k = P_3+P_4+P_5+P_6+P_7+P_8$$ We can continue these iterations until we have a string of $2$s and $3$s.

For example: $$24 = 13 + 11$$ $$24 = (2+2+2+2+2+3)+(2+3+3+3)$$

This is the same as the Diophantine equation: $$24 = 2a + 3b$$ We can group together $2$s and $3$s in certain configurations.

Instead of $$24 = (2+2+2+2+2+3)+(2+3+3+3)$$ we could have: $$24 = (2+3)+(2+3)+(2+2+3)+(2+3+3)$$ Which is equal to: $$24 = 5c + 7d$$ Giving us the general: $$2k = 5c +7d$$ However, these Diophantine equations also represent the primes. $$P_x = 5c + 7d$$ Where $7\not|c$ and $5\not|d$.

While finding solutions for these for prime numbers I found that some primes required negative solutions for $c$ and $d$. I called these the exceptions.

Examples of these exceptions are $11, 13, 23$. However, here I found the exceptions stopped. I checked the Diophantine equations for other primes, up to 179 and they all had solutions.

We can prove $23$ is the greatest prime to require negative numbers, by observing that if we had our target prime $P_x$ and a number which we already know has solutions then we can add to that number to approach our prime.

First, we could add $2$, by subtracting $5$ and adding $7$. This means we can repeat this to add $4,6,8,10$ etc. We can also add $1$ by subtracting $5(4) = 20$ and adding $7(3) = 21, 21-20 =1$.

But because this requires us to subtract $20$ we must be able to form $P_x$ using $4$ or more $5$s. Since $23$ is the next biggest prime above $20$, it is the last one to be formed with less than $4$ $5$s.

If we take an example, $$ 39 = 5(5) + 7(2)$$ $$ 41 = 5(4) + 7(3)$$ $$ 43 = 5(3) + 7(3)$$

Here we have produced $41$ and $43$ from $39$.

This means that any prime, excluding the exceptions, can be represented as: $$P_x = 5c+7d$$

If we now go back to: $$2k = P_1+P_2$$ The next even number is $2k +2$. This is equivalent to adding $2$ to $5c+7d$. We can then rearrange this into different blocks of $5$s and $7$s and add them up into primes as we have shown to be possible. This means: $$2k + 2 = P_1+P_2$$ Therefore, every even number $>2$ can be written as the sum of two primes.

Therefore, the Goldbach conjecture is True.

Q.E.D.

Can anyone point out my error, as an elementary "proof", such as this should simply not be possible.

Thank you.

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  • $\begingroup$ The easiest problem to recognize is that if $c$ and $d$ have a common factor, then $5c+7d$ will be composite, and you have not shown that the groupings can always be chosen such that $\gcd{c,d}=1$. $\endgroup$ – Keith Backman Aug 4 '18 at 20:57
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We can then rearrange this into different blocks of $5$s and $7$s and add them up into primes as we have shown to be possible.

You have shown no such thing. You have shown that given any prime (greater than $23$), you can make it out of blocks of $5$s and $7$s. This in no way says that given any collection of $5$s and $7$s, you can group them into two parts where each part adds up to a prime.

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