Question

Infinite twin primes (minus some group theory or something)

Answer 1

The "twin primes constant" is, as the name suggests, constant.  It is calculated using a (high-level) similar method to your cake analogy.  This fact confirms neither infinite nor finite status on the set of twin prime pairs.

Answer 2

A good chunk of what you wrote has certainly already been formalized and proven.

I love the visual of cutting a cake and seeing how much is left as a portion of the whole. This line of thinking leads one to the idea of "natural density", in roughly the same terms you gave, but with the help of limits.

But natural density is kind of broad and you need lots of hedges to squeeze out something useful when it comes to slicing up slices. Luckily for you, you have noticed one major hedge: the candidates form a cyclical pattern.

And this is where number theory (where you should start if you are interested in primes) comes in. We call a set of numbers a "reduced residue system modulo n" *exactly* in the situation you have described: it's what's left over in one cycle of length n after you take out all the stuff that isn't coprime to n. So from 7 to 12,  the set {7,11} forms a reduced residue system modulo 6.

A residue class is what you get when you consider all the "equivalent" candidates (for example, in your k=2 example, 7 and 13). The nicest thing about residue classes is that they form arithmetic progressions.

And that's the ingredient that you'll need to make the density/slicing argument work, essentially that the "slice" is evenly distributed among the whole cake and won't be intefered with by past slices. Your infinitude of primes "proof" is good.

Now I'm no expert in number theory or anything like that, but I think where your idea starts to fall apart is that when you slice out primes, each prime affects only its own kind, but with twin primes the removal of one prime can knock out up to two others: for example 13-15-17 is *two* sets of twin prime candidates, but knocking out 15 knocks out both (13,15) and (15,17) from candidacy. (I know the example isn't perfect, cuz 11-13 still holds, but hopefully you get the idea).

Answer 3

Let's focus on your argument for the infinitude of primes. The usual sieve looks like this: at the k-th step, remove all natural numbers divisible by p_k, *except for p_k itself*. What if I also remove p_k at the k-th step? Then there's nothing left at the end of the sieve, even though the density of "prime candidates" in the next cycle (from p_(k+1) to p_1p_2...p_k+p_(k+1)) remains the same. Your density argument is not sensitive to removing one more number at each step, but the outcome (nothing remaining instead of infinitely many primes) is drastically different.

Answer 4

For your proof that there are infinite primes, the key assertion can be proven using the Chinese remainder theorem, which states that numbers from (P1...Pk) to (2P1...PK-1) are in one-to-one correspondence with lists of possible remainders upon dividing by P1,...,Pk. (For example if k=2 then 11 corresponds to the list of remainders (1,2) ). And its clear that exactly 1/k of the lists of remainders that have no zeroes in the first k-1 entries will have a zero in the kth entry.

But the similar assertion for twin primes is almost certainly extremely difficult to prove.