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I‘m trying to prove the following

Conjecture. There exist an infinite number of positive integers $n$ not of the form $$n=\frac{u^2-u+v^2+3v}{4(u-v)},$$ where $u$ and $v$ are positive integers.

Brute force computation seems to support my claim, e.g., for $1 \le u,v \le 1000$, the form does not represent $${1,2,3,6,13,16,18,22,23,\dots}$$

I’ve done some basic algebraic manipulations, but haven’t found the magic incantation. Any help or hints would be appreciated.

EDIT: It appears that even the form of the numerator ($u^2-u+v^2+3v$) may not be “sufficiently universal”. A modified conjecture to that effect might be easier to prove, and would evidently serve my purposes just fine (since if the numerator does not represent all positive integer multiples of $4$, then the fraction clearly cannot represent all positive integers).

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THIS IS A PARTIAL ANSWER

Only an approach, but too long for a comment.

The equation $$n=\frac{u^2-u+v^2+3v}{4(u-v)}$$ is equivalent to $$u(4n-u+1)=v(4n+v+3)$$

So, for every $n$ , there are only finite possible numbers $u$ because the right side is positive and the left side is positive if and only if $u<4n+1$

$n=1$ can , for example, be ruled out, since $u(5-u)$ can only have the values $4$ and $6$ , but the right side is clearly at least $8$

At least, this approach allows to verify whether a special value occurs , in a finite number of trials. No idea whether infinite many $n$ are missing.

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