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Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y \in \Bbb Z$ ?

I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y \in \Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|(x^{2} + y^{2} + 1) $. : the sets $\{-1-x^2 \mid x \in \Bbb F_p\}$ and $\{y^2 \mid y \in \Bbb F_p\}$ have both cardinality $(p+1)/2$, so they must have a non-empty intersection, i.e. an element of the form $z=-1-x^2=y^2$.

But the proof does not work to show that all but finitely many (or at least infinitely many) primes divide $x^2+xy+y^2+1$ for some integers $x,y$. So what can I do?

Alphonse
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If $p\ne2$ then $p\mid(x^2+xy+y^2+1)$ iff $p\mid(4x^2+4xy+4y^2+4)$ that is iff $p\mid((2x+y)^2+3y^2+4)$. The same trick as before proves that there are $y$ and $z$ with $p\mid(z^2+3y^2+4)$.

Angina Seng
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  • Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$... – Alphonse Dec 02 '18 at 12:27
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    @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)\equiv0$. – Angina Seng Dec 02 '18 at 12:33