Evidence against Goldbach's Conjecture?

It recently occurred to me that, unless I'm much mistaken, Goldbach's conjecture can easily be seen to be equivalent to a seemingly more general statement:

Every number $n$ divisible by any $1<d<n$ is sum of $d$ primes. (And so every number $>1$ would be average of an arbitrary multitude of primes?!)

Proof

GC is the special case when $d=2$. Conversely, use strong induction on $d$: If $d=2$, then this is GC. Suppose $d=3$. Then $n\geq 6$. If $n$ is even, then $n-2 (\geq 4)$ is sum of $2$ primes by GC. If $n$ is odd, then $n-3 (\geq 4)$ is sum of $2$ primes by GC. So in any case $n$ is sum of $3$ primes. Suppose now $d\geq 4$ and suppose result holds true for all divisors $<d$. Write $d=d_{1}+d_{2}$ with $d_{1}, d_{2}\geq 2$. Then $n=dr=d_{1}r+d_{2}r$ with $r>1$. So by strong induction $d_{1}r$ is sum of $d_{1}$ and $d_{2}r$ is sum of $d_{2}$ primes. Hence $n$ is sum of $d_{1}+d_{2}=d$ primes.

As far as I know no one seriously doubts that GC is true. So why is the above no legitimate reason to believe that GC is false? The generalized statement seems ridiculously strong to me...

Solution 1:

Essentially, because your generalization doesn't add any strength. This already follows from the fact that it's derivable from Goldbach, as you've shown, but there's also a conceptual reason: as $d$ goes up the strength of the statement for that specific $d$ goes down drastically. In fact, we already know a version of the first step along your generalization: every sufficiently large odd number is the sum of three primes! Conceptually, having more primes available (even if the number is fixed) gives you many, many more options for building numbers out of them.

One way you might see this is to study the analagous problem using squares instead of prime numbers. Obviously, the number of numbers that are 'sums' of $1$ square is small - it's only the squares (so in particular there are $\Theta(\sqrt{N})$ of them $\leq N$). On the other hand, every prime $p=4n+1$ (and products of these primes, and the products of those by squares) is the sum of two squares; numbers that are the sums of two squares still have density zero, but there are $\Theta(\frac{N}{\sqrt{\log N}})$ of them $\leq N$. If we add another number in, then most numbers (approximately $\frac56N$) $\leq N$ are the sum of three squares — and of course, every number is the sum of four squares.

Solution 2:

The problem is in "The generalized statement seems ridiculously strong to me...". While it's true that you're adding conditions it's also not so difficult to notice that you're not indeed adding any real strength.

To see the problem more clearly consider the absolutely trivial:

• Any non-negative number $n$ is the sum of 4 squares
• Any non-negative number $n$ is the sum of $k$ squares for each $k \ge 4$

The second statement is "stronger" (formally) because includes the first as a subcase, and also adds an infinity of conditions. But does it really add any strength? Of course not because 0 is a perfect square and all the real work is in the case for 4.

This happens because the added conditions are weaker and weaker as $k$ increases and the same happens increasing $d$ in your case.

Everything simply weights on the base case and other stuff is just smoke (trivial to see in this case, just slightly less trivial to see in your case).