What numbers are not represented by $5xy+2x+2y$? Do they have a positive density?


This came up for me while investigating some cases [here](https://mathoverflow.net/questions/338050/is-multilinear-hilberts-tenth-problem-version-undecidable/339014?noredirect=1#comment848132_339014). Here's what I've found:


* All evens are represented with $x=0$, and all $3m+1$ are represented with $x=-1$.
* There are infinitely many $n$ not represented, e.g. any $n$ for which $5n+4$ is prime.
* If $5xy+2x+2y=n$, then either $|x|$ or $|y|$ is less than $|n|/5+2$, so for each $n$ this is decidable.
* Of numbers with absolute value less than 6000, about 80% are represented by this polynomial.


I'd expect a nice characterization for these numbers, but I haven't found it.



What numbers are not represented by $5xy+2x+2y$?
$n = 5xy + 2x + 2y$ if and only if $5n+4 = (5x+2)(5y+2)$.
So a necessary and sufficient condition is that $5n+4$ have
a factor congruent to $2 \bmod 5$ --- or $3 \bmod 5$ since you're
allowing negative $x,y$ such as $x = -1$. This makes it easy to decide
whether a given $n$ is so represented. In particular, the numbers 
*not* represented by $5xy + 2x + 2y$ are those for which $5n+4$ is 
the product of primes all congruent to $\pm 1 \bmod 5$ 
(you already found the special case of prime $5n+4$).
Such numbers have density zero, but convergence is slow:
the density in $|n| < X$ is asymptotically proportional to $1 \left / \sqrt{\log X} \right.$.


P.S. Here's some quick **gp** code to count such $n$ up to $N$:



```
f(v) = sum(n=1,#v,(v[n]%5!=1)&&(v[n]%5!=4))
F(n) = f(factor(n)[,1])==0
S(N) = sum(n=1,N,F(5*n+4))

```

For $N=6000$ the count $S(N)$ is $1204$, which agrees with **Matt F.**'s
calculation that "about 80%" of $n \leq 6000$ *are* represented by $5xy+2x+2y$.
Taking $N=10^5, 10^6, 10^7, 10^8$ finds $S(N) = 17992$, $166612$, $1557892$, $14680787$
(the last count took about 5.5 minutes to compute); this is quite close to
$CN \left / \sqrt{\log(5N)} \right.$
for $C$ somewhere between $0.65$ and $0.66\,$.


