Consider Z+ and the function |6xy+x+y| : x,y Z \ {0} within Z+.
Is there an asymptotic relationship between the integers in Z+ and |6xy+x+y|, where |6xy+x+y| can never fill the whole set of integers?
If this is asymptotic and can never meet the asymptote (defined by n=|6k+1| for n=|36xy+6x+6y+1|) ; does this mean that there would be infinite positive integers of the form k not expressible as k=|6xy+x+y| since this is a Diophantine equation?
Here are two very similar provable examples which do not involve absolute value within Z+.
- n=k+1 \ n=xy+x+y+1 is all prime numbers (therefore if k=xy+x+y, then n=k+1 is composite).
- n=2k+1 \ n=4xy+2x+2y+1 is all odd primes (therefore if k=2xy+x+y, then n=2k+1 is composite).
When we move to an absolute value expression, then we have to move to Z \ {0} for some variables, and we need to focus on k as opposed to n when identifying compositeness.
So we have n=6k+-1 numbers. Then we have n=6k-1 and n=6k+1.
But {|6k-1|}={|6k+1|} so we can just use n=|6k+1| if n is Z+ and k is Z{0} (or choose n=6k-1 if you want but it is less pedagogically sound as it will flip signs in next steps).
Then we parameterize from n=|36xy+6x+6y+1| (for n in Z+, x,y in Z \ {0}), reducing to k=6xy+x+y (k,x,y in Z \ {0}).
So, if -k is in k=6xy+x+y, then n=|6k+1| yields a composite number of the form n=6k-1.
If +k is k=6xy+x+y, then n=|6k+1| yields a composite number of the form n=6k+1.
By Dirichlet’s theorem there are infinite primes in n=6k-1 and n=6k+1, so there are infinite k values yielding prime numbers in n=|6k+1|.
(A more formal proof demonstrates complete coverage for composites of n=6k+-1 integers by this method, so it is not necessary to type out here, but don’t forget {|6k-1|}={|6k+1|} makes this all possible, for every n in one there is -n in the other and vice versa.)
In short, if k \ 6xy+x+y for k,x,y in Z \ {0}, then k yields a prime in n=6k-1 or n=6k+1 when using n=|6k+1| as our arithmetic progression.
Since there are infinite primes, then this covers every prime in n=6k+-1 and so covers all primes greater than 3.
All of these, and due to the proof of the infinitude of composites and primes in each sequence ought to demonstrate an asymptotic relationship.
Since these are Diophantine sets with only integer solutions, does this mean in each case that the asymptotic relationship with the arithmetic progression which generates the composite parameterization guarantees there will be infinite integers not expressible of the form of the originating arithmetic progression?
It’s like the fundamental theorem of arithmetic applied to arithmetic progressions.
Imagine that k is a line, and k=|6xy+x+y| is curved (e.g. asymptotic with k), and so it can never hit the line even if it fills almost all of k when k is very large. Therefore, it is logically necessary that the complement of k=|6xy+x+y| is an infinite set of integers since the equation produces only integer solutions.
We just need to establish a baseline of an arithmetic progression which becomes the “asymptotic value”. The simplest example is the set of all primes n=k+1 \ n=xy+x+y+1, which provably never touches all the integers.
Since the parameterized sequences in all these examples are Diophantine and infinite in Z+ but cannot possibly become a “line” by transforming into their parent arithmetic progression, then it follows logically that there are infinite positive integers k not expressible as |6xy+x+y| as well. You would be trying to change the dimensional relationships in a way which is not possible.
Arguing the opposite is absurd.
Go ahead and show me how k in Z+ can become |6xy+x+y| : x,y Z \{0} just because k is big. If you insist at some magical point that |6xy+x+y| will become infinitely co-finite in Z+, you will be showing me you are trying at hammering a curve into a line, and it will never be a line. Enjoy counting sand.
Hand written mo fo (but taught by AI over a year, lol).