4
$\begingroup$

It is relatively easy to prove that there exist infinitely many positive integers not of the form $2ij+i+j$, where $i,j\in \mathbb N$.

Sundaram already found that some positive integer is prime if and only if it is of the form $2k+1$, were $k$ is some positive integer not of the form $2ij+i+j$. The proof is pretty straightforward, as $2(2ij+i+j)+1=(2i+1)(2j+1)$. And as we have many proofs of the existence of infinitely many prime numbers, we can use any of them together with Sundaram's result to prove the existence of infinitely many positive integers not of the form $2ij+i+j$.

My question

Do you know of / can you share any proof of the existence of infinitely many positive integers not of the form $2ij+i+j$ that does not use neither Sundaram's result nor some classical proofs of the existence of infinitely many prime numbers? For instance, a proof using sieve theory would be acceptable.

Thanks in advance!

EDIT

As a clarification, I am not looking for alternative proofs that there are infinitely many primes. I am looking for a proof that there are infinitely many positive integers not of the form $2ij+i+j$, because maybe it can be useful to derive insights for the generalized form $Kij+i+j$, $K\in\mathbb N$. Incidentally, the fact that there are infinitely many primes tell us (through Sundaram's) that there are infinitely many positive integers not of the form $2ij+i+j$, but I have not been able to extrapolate something useful for the generalized problem.

$\endgroup$
13
  • 1
    $\begingroup$ @Snared What do you mean? The least number that is so expressible is $F(1,1)=4$, so $1,2,3$ can't be so written. Then $F(2,1)=F(1,2)$ is next and that's $7$, so neither $5$ nor $6$ is so expressible. Indeed, as the OP argues, $k$ is inexpressible iff $2k+1$ is prime. Of course, we are asked to not use the equivalence is arguing the infinitide. $\endgroup$
    – lulu
    Aug 10, 2023 at 20:05
  • 1
    $\begingroup$ yup I messed up my script, I see now. I had j=0 so was getting everything back $\endgroup$
    – Snared
    Aug 10, 2023 at 20:10
  • 1
    $\begingroup$ Usual problem: in some countries (including mine), $\mathbb{N}$ includes $0$, and "positive" includes $0$ too. $\endgroup$ Aug 10, 2023 at 20:12
  • 3
    $\begingroup$ This is essentially a sieve issue. Fixing $i=1$ gives the sequences $\{4, 7, 10, 13, \cdots\}$, i.e. $n\equiv 1 \pmod 3$, starting at $4$. Similarly, $i=2$ gives $n\equiv 2\pmod 5$, starting at $7$, and so on. Look at the progressions generated this way and try to argue that the first term grows too fast to allow the progressions to cover everything. The starting terms are $\{4, 7, 10, 13, \cdots\}$ so, again, the integers $n\equiv 1 \pmod 3$ starting at $4$. Not sure this will work without knowing that there are infinitely many primes, though. $\endgroup$
    – lulu
    Aug 10, 2023 at 20:22
  • 6
    $\begingroup$ It is unclear what sort of proof would be acceptable to OP. What would it mean to not use the existence of infinitely many prime numbers, to prove something that is plainly equivalent to the existence of infinitely many (odd) prime numbers? $\endgroup$
    – Erick Wong
    Aug 10, 2023 at 20:22

0

You must log in to answer this question.