Today we introduced Goldbach's Conjecture and how it has been modified and verified to varying degrees since its inception in 1742. This includes the strong and weak versions of the conjecture. After this we discussed the different mathematicians who had worked on the conjecture and their work. Lastly, we explained through a heuristic argument why you should believe Goldbach's Conjecture to be true.

Let's Begin

On June 7, 1742, Christian Goldbach, a Prussian mathematician, wrote a letter to Leohnard Euler proposing a conjecture. Through their correspondence, the following conjecture came about: Every integer greater than 2 can be written as the sum of three primes.
Note: At this time, 1 was considered to be a prime number.


  • $4 = 1 + 1 + 2$
  • $20 = 11+ 7 + 2$

Goldbach's Strong and Weak Conjectures

Since 1 is no longer considered to be prime, we had to amend Goldbach's original proposal to what is known today as:

Goldbach's Weak Conjecture:

All odd numbers greater than 7 can be written as the sum of three odd primes.

This was changed because since 1 is no longer considered prime, we cannot write any number less than 7 as the sum of three odd primes. For example, $5 = 1+ 2 + 2$ needs 1 in order to be written as the sum of three "primes". 7 is not a possibility either since we need three odd primes, and $7 = 3 + 2 + 2$, and 2 is not an odd prime.

In order to prove the weak conjecture, mathematicians came up with a stronger result:

Goldbach's Strong Conjecture:

Every even integer greater than 2 can be written as the sum of two primes.

Note: We have seen this before, in Chapter 1, page 15.

Does the strong imply the weak?

If we can prove the strong conjecture to be true, we need only add some fixed odd prime to get an odd integer as the sum of three odd primes. For example:
$28 = 11 + 17 \Rightarrow 31 = 11 + 17 + 3$. We do not need to add 3 every time. It could be any odd prime number, and we will directly get an expression that holds true under Goldbach's Weak Conjecture.

Does the weak imply the strong?

Along the same lines as above, if the weak implied the strong, we could subtract any odd prime from our expression and get something that holds true under Goldbach's Strong Conjecture. Let's look at 31. $31 = 11 + 13 + 7 \Rightarrow 28 = 11 + 13 + 4$ which is not the sum of two odd prime numbers. Therefore, since we cannot directly get a way to express an even integer as the sum of two odd primes by subtracting any odd prime, the weak does not imply the strong.

A Little Bit Of History

First, we need to time warp back in time.

$\bigstar$ N. Pipping (1938)

This was the last time the Cubs won 100 games in a season, and fortunately for the math community, N. Pipping was not enjoying great baseball, as he instead:

Directly verified that Goldbach's Strong Conjecture works for all numbers $n \leq 10^{5}$.

Note: Since this works for the strong, it will also work for the weak.

Now it's time for another time warp!

$\bigstar$ Lev Schnirelmann (1930)

While the Dust Bowl made everything dry, mathematics was not because Schnirelmann proved:

Every even $n \geq 4$ can be written as the sum of at most 300,000 primes.


  • $12 = 3 + 5 + 2 + 2$, which is less than 300,000 primes
  • $12 = 2 + 2 + 2 + 2 + 2 + 2$, which is again less than 300,000 primes.

You guys know what time it is! Time warp!

$\bigstar$ Olivier Ramare (1995)

While OJ was on trial, Schnirelmann's work was on trial in the mathematics community. As a result, Ramare came up with this defense:

Every even number $n \geq 4$ is in fact the sum of at most six primes.


  • $12 = 2 + 2 + 2 + 2 + 2 + 2$ which is less than or equal to 6!
  • $14 = 2 + 2 + 2 + 2 + 2 + 2 + 2$, which is not less than 6. However, Ramare's work explains that this is not that only way, and there is a way to write it in six or less, i.e. $14 = 3 + 3 + 2 + 2 + 2 + 2 = 3 + 5 + 3 + 3$, both of which are less than or equal to 6.

A good way to think of this is in terms of a rope and tying a knot. Schnirelmann took a lot of rope with a lot of slack to tie a knot. Ramare felt this was too much slack, and tied his knot with a lot less rope.

Before we move on to our next mathematician, we need to introduce another definition.

Sufficiently large

There exists an N such that for all $n \geq N$ some statement holds true.

This is important because it provides a lower bound for numbers that do work for a given statement.

Ok. Time warp!

$\bigstar$ Ivan Vinogradov (1937)

During this time the Great Depression was still raging on, but the mathematics world has a reason to not be depressed, since Vinogradov came up with the following:

All sufficiently large odd numbers n can be written as the sum of three primes.

Basically this says that Goldbach's Weak Conjecture holds for all but a finite amount of odd numbers.

You can think of this statement as a net that Vinogradov casts out into the sea of numbers. All numbers caught in the net are "sufficiently large" and work. Numbers that aren't caught don't necessarily not work, we just don't know. If this is confusing, ask any of the groups members for clarification.

We're done in 1937, so let's do another time warp!

$\bigstar$ Hyman Levy (1963)

Beatlemania was reaching it's peak, Levymania in the math world had reached its peak because he conjectured:

Levy's Conjecture, aka the "Medium Strength" Conjecture

Every odd number greater than or equal to 7 can be written as the sum of a prime and twice a prime.


  • $13 = 3 + (2 * 5)$
  • $31 = 5 + (2 * 13) = 17 + (2 * 7)$

Again, we need to introduce another concept before we move on to our next mathematician.


The product of two primes. I.e., $77 = 7 * 11$.

Note: Semiprimes were used in Levy's Conjecture.

Now we can time warp to our next guy!

$\bigstar$ Chen Jingrun (1973)

Math was getting funky along with the rest of the world because this far out guy proved:

Every sufficiently large even number can be written as either the sum of two primes or a prime and a semiprime.


  • $16 = 2 + (2 * 7)$
  • $100 = 23 + (7 * 11)$

This is similar to Levy's Conjecture, but it is more powerful.

Time for our last spin into the past. Time warp!

$\bigstar$ Heath-Brown and Schlage-Puchta (2002)

While Kelly Clarkson won the hearts of America on American Idol, these guys were winning the hearts of mathematicians everywhere for proving:

Every sufficiently large even integer is the sum of two primes and exactly thirteen powers of 2.

Note: This to our knowledge is the most recent work on Goldbach's Conjecture.

Heuristic Justification

This provides an explanation of why you should believe Goldbach's should be true, since obviously there is no proof yet.

A key theorem used is the Prime Number Theorem.

Prime Number Theorem

The limit as $x \rightarrow \infty$, $\frac{\pi(x)}{\frac{\(x}{\ln(x)}} = 1$, where $\pi(x)$ counts the number of primes less than or equal to x.
In other words, as x gets really really big $\pi(x) \sim \frac{\(x}{\ln(x)}$.

Therefore, the probability a number n is prime $\frac{1}{\ln(n)}$, since the total number of primes divided by the total amount of numbers is $\frac{\frac{\(n}{\ln(n)}}{\(n} = \frac{1}{\ln(n)}$.

So if we want to write some number n as the sum of two numbers, we can do $n = (n - m) + m$. The probability that $(n - m)$ and $m$ are prime is $\frac{1}{\ln(n - m)}$ and $\frac{1}{\ln(m)}$. Assuming independence, the probability both are primes is $\frac{1}{\ln(n - m)} * \frac{1}{\ln(m)}$.

When we add these probabilities up we would get $\sum_{m = 1} ^ {\frac{n}{2}} \frac{1}{\ln(n - m)} * \frac{1}{\ln(m)}$. We cannot figure out this sum, but we can get a lower bound for it.

We know that if $n \geq m$ then $\frac{1}{n} \leq \frac{1}{m}$. Therefore, $\frac{1}{\ln(n)} \leq \frac{1}{\ln(n - m)}$ and $\frac{1}{\ln(m)}$. So we can say that $\sum_{m = 1} ^ {\frac{n}{2}} \frac{1}{\ln(n - m)} * \frac{1}{\ln(m)} \geq \sum_{m = 1} ^ {\frac{n}{2}} \frac{1}{\ln(n)} * \frac{1}{\ln(n)} = \frac{\(n}{2 * \ln(n) ^ {2}}$.

Since $n$ is a much stronger function than $\ln(n)$, $\frac{\(n}{2 * \ln(n) ^ {2}}$ gets really big really quickly. Thus since our sum is larger than a really big number, our sum is even bigger. This is because the number of times you try to find two primes that add up to n outweighs the probability those numbers added are prime. Therefore, it makes sense that eventually, we will find two prime numbers whose sum is n.

Now, this is not the best argument since it assumes $(n - m)$ and $m$ are independent of each other, which makes our probability both are prime higher than it would be.

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