A cubic equation, and integers of the form $a^2+32b^2$

Question

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)?

Answer 1

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is $\pm 1$ modulo 8, $2 \mid \nu_q(p)$ for any prime $q \equiv 7 \pmod{8}$. Then $p$ has an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q^{\nu_q(p)}}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Will Jagy showed that the representation $$ p = (x^2 + 4)^2 - 2(2x + 4)^2 $$ is primitive and $p$ is not divisible by any prime $q \equiv \pm 3 \pmod{8}$. Then from $p = u^2 + 32v^2$ we see that $d = \gcd(u, v) \equiv \pm 1 \pmod{8}$, $p = p_1d^2$, $p_1$ has only divisors of the form $8k + 1$. $u \equiv \pm 1 \pmod{8}$ since $p \equiv 1 \pmod{16}$. $$p_1 = \left(\frac{u}{d}\right)^2 + 32 \left(\frac{v}{d}\right)^2$$ Applying theorem 1 we obtain that $p$ is divisible by an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power.

$$p_1d^2 = (x^2 + 4)^2 - 2(2x + 4)^2$$ Applying theorem 2 we obtain $$\left(\frac{(x^2 + 4)(2x + 4)}{p_1}\right) = \left(\frac{(x^2 + 4)(2x + 4)}{p_1d^2}\right) = \left(\frac{-2}{x^2 + 4}\right) = -1 $$ Therefore $p$ is divisible by an odd number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power. And this is a contradiction.

Answer 2

details, details. From $x$ odd and

$$ x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2. $$ we see that $x^4-32x-16$ is not divisible by any prime $q \equiv 3,5 \pmod 8.$ That is, $x^2 + 4$ is also odd. Next, $\gcd(x^2 + 4, 2x+4) =\gcd(x^2 + 4, x+2).$ However, $(x+2)(x-2) = x^2 - 4,$ so that $\gcd(x^2 + 4, x+2)$ divides $8$ and must be $1.$ Thus the representation $x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2$ is primitive.

Now, $x^4 - 32 x - 16$ is often divisible by primes $7 \pmod 8$ (with odd exponent) and are therefore ruled out. Next, it appears that $x^4 - 32 x - 16$ has always an odd number of prime factors of shape $4 u^2 + 4uv + 9 v^2,$ and I am fiddling with showing that these numbers are a dead end as well. This could require understanding every word in Liu and Williams, hard to say

Answer 3

Hm, unless I am mistaken, there is a much shorter proof than presented by Denis. It is clear that $y$ is negative and odd. Multiply the equation by $y$ and rewrite as $$ y-1+(1+xy)^2+4y^3+2yz^2=0. $$ Now denote $X=1+xy$, then do substitution $y \to -y$, and rewrite the equation as $$ X^2-2yz^2 = 4y^3+y+1 = (2y+1)(2y^2-y+1), $$ where $y$ is now positive and odd. Consider the equation as quadratic in $X,z$ with parameter $y$. The discriminant of the left-hand side is $D=8y$, hence the equation implies that the Jacoby symbol $\left(\frac{8 y}{2 y+1}\right)$ is equal to $1$. However, computation shows that it is equal to $-1$ for every odd positive $y$.

Answer 4

The equation $(1)$ has not integer solution. An elementary proof.

Suppose $(x,y,z)=(a,b,c)$ is a solution and consider as unknowns $(X,Y)=(4,2)$ so we can form the system $$\begin{cases}b^2X+(a+c^2)Y=-(1+a^2b)\\-X+7Y=10\end{cases}$$ whose solution is $$X=\dfrac{-7-7a^2b-10a-10c^2}{7b^2-a-c^2}$$ $$Y=\dfrac{10b^2+1+a^2b}{7b^2-a-c^2}$$ Because of $X=2Y$, one has after simplification $$9(1+a^2b)+10(a+c^2+2b^2)=0$$ and multiplying equation $1+2a+a^2b+4b^2+2c^2=0$ by $5$ we have $$5(1+a^2b)+10(a+c^2+2b^2)=0$$ Subtraction now gives $$4(1+a^2b)=0$$ We are done.