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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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    $\begingroup$ journals.math.tku.edu.tw/index.php/TKJM/article/view/4461/1510 $\endgroup$
    – Will Jagy
    Dec 6, 2023 at 17:05
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    $\begingroup$ the rest of the primes $1 \pmod 8$ are represented by $4x^2 + 4xy + 9 y^2 $ $\endgroup$
    – Will Jagy
    Dec 6, 2023 at 17:06
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    $\begingroup$ sorry, your prime must have roots of $u^4 - 2 u^2 + 2,$ or $(u^2 - 1)^2 \equiv -1 \pmod p.$ This is from Liu and Williams (1994) Tamkang J., link above $\endgroup$
    – Will Jagy
    Dec 6, 2023 at 17:13
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    $\begingroup$ Interesting. If this statement (either representation $s^2+32z^2$ or $4x^2+4xy+9y^2$ is true but not both) remained correct for all integers equal to $1$ mod $4$, then it would be left to represent $P(t)$ as $4x^2+4xy+9y^2$ for all $t$. But this is not true for composites: integer $33$ is representable in both ways. $\endgroup$ Dec 6, 2023 at 17:16
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    $\begingroup$ Right. The other two forms (classes) are $3 x^2 \pm 2xy + 11 y^2,$ these forming the non-principal genus. These are more predictable, both represent all primes $3 \pmod 8.$ The square of one of these by Gauss composition is $\langle 4,4,9 \rangle, $ but they are "opposites" and composition takes that pair to the identity $\langle 1,0,32 \rangle, $ Actually, you get both ways for any numberr $1 \pmod 8$ once there are no prime factors $5,7 \pmod 8$ $\endgroup$
    – Will Jagy
    Dec 6, 2023 at 17:28

4 Answers 4

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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.

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    $\begingroup$ Wow, thank you! The proof would be easier to follow if you provide more details in places such as "similar analysis", "modulo p considerations shows", "can be generalized", etc. $\endgroup$ Dec 15, 2023 at 19:00
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    $\begingroup$ @BogdanGrechuk The proof definitely needs more explanation. What I initially missed is that the $p = u^2 + 32v^2$ representation is not necessarily primitive. I will rewrite the proof tomorrow. $\endgroup$ Dec 15, 2023 at 21:04
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    $\begingroup$ Thank you, this presentation is much better, but I still do not understand whether you applied Theorem 2 to $p'$ or to $p$. For $p'$, I do not see how you get representation $p'=a^2-8b^2$. If you apply it to $p$, how you get a condition that every prime divisor of $p$ is $1$ modulo $8$? $p=p_1 d^2$, and $d$ can have prime factors equal to $-1$ modulo $8$. $\endgroup$ Dec 17, 2023 at 20:08
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    $\begingroup$ Ok, I think I found the way around this. We should relax the condition that every prime divisor of p is 1 modulo 8 to the condition that every prime divisor of p that enters its prime factorization in an odd exponent is 1 modulo 8. Other odd prime factors are allowed to be arbitrary. The proof of Theorems works for this more general statement as well, and it suffices for application to our equation. $\endgroup$ Dec 17, 2023 at 21:44
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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

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    $\begingroup$ it appears that $x^4 - 32 x - 16$ has always an odd number of prime factors of shape $4 u^2 + 4uv + 9 v^2$ - why? $\endgroup$ Dec 7, 2023 at 2:23
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    $\begingroup$ @Fedor right. Can't prove it yet. I just wrote a little program to check odd $x$ up to some bound. $\endgroup$
    – Will Jagy
    Dec 7, 2023 at 2:32
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    $\begingroup$ You write that the case when $x^4-32x-16$ is divisible by some primes equal to $7$ modulo $8$ is ruled out. Why? Such primes can be common factors of $s$ and $z$.... $\endgroup$ Dec 17, 2023 at 20:10
  • $\begingroup$ @BogdanGrechuk I was a bit careless. $\endgroup$
    – Will Jagy
    Dec 17, 2023 at 20:51
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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 $y$.

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  • $\begingroup$ $8y\equiv-4\bmod{2y+1}$. When $y$ is odd, $2y+1\equiv3\bmod4$. $\endgroup$ Feb 17 at 22:30
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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.

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  • $\begingroup$ Unfortunately this doesn't work; after correcting a couple of sign errors, the third-to-last displayed formula comes out the same as the second-to-last, and the final subtraction merely yields $0=0$. $\endgroup$
    – GNiklasch
    Dec 25, 2023 at 15:33
  • $\begingroup$ I made frecuently mistakes in calculation. Thanks you. $\endgroup$
    – Piquito
    Dec 25, 2023 at 17:14
  • $\begingroup$ @GNiklasch.- Thanks you very much really. For my age I should retire from this now. $\endgroup$
    – Piquito
    Dec 25, 2023 at 17:40

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