I am trying to determine whether there are any integers $x,y,z$ such that $$ 1+2 x+x^2 y+4 y^2+2 z^2 = 0. \quad\quad\quad (1) $$ It is clear that $x$ is odd. We can consider this equation as quadratic in $(y,z)$ with parameter $x$. After multiplying by $16$, we can rewrite the equation as $$ (8y+x^2)^2+32z^2=x^4-32x-16. $$ Writing $x=2t+1$ and denoting $s=8y+x^2$, we obtain $$ s^2+32z^2=P(t), \quad\quad\quad (2) $$ where $P(t)=(2t+1)^4-32(2t+1)-16$. If this equation has no integer solutions, then so is the original one.

With $Z=2z$ we can rewrite it as $$ s^2+8Z^2=P(t) \quad\quad\quad (3) $$ with $Z$ even. It is known that every prime congruent to $1$ modulo $8$ is of the form $s^2+8Z^2$. $P(t)$ is always equal to $1$ modulo $8$, and takes infinitely many prime values by Bunyakovsky conjecture. By finding $t$ such that $P(t)$ is prime, we can generate as many solutions to (3) as we want, but $Z$ happens to be always odd in all these solutions up to a large bound.

Unfortunately, there seems to be no coprime $a,b$ such that all primes equal to $a$ modulo $b$ are of the form $s^2+32z^2$, so we cannot easily apply the same method directly to (2).

So, are there any integers $x,y,z$ satisfying (1)?

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