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The question is: are there any integers $x,y,z$ such that $$ 1+4 x^3+x y^2+2 y z^2 = 0 \quad\quad\quad\quad (1) $$

The motivation is: Define the length of a polynomial $P$ consisting of $k$ monomials of degrees $d_1,\dots,d_k$ and integer coefficients $a_1,...,a_k$ as $l(P)=\sum_{i=1}^k\log_2|a_i|+\sum_{i=1}^k d_i$. This is an approximation for the number of symbols used to write down $P$ if we write the coefficients in binary, do not use the power symbol, and do not count the operations symbols.

If we study the solvability of Diophantine equations ordered by length, then the smallest open ones are $ y(x^3-y) = z^4+1, $ $ 2 y^3 + x y + x^4 + 1 = 0 $ and $ x^3 y^2 = z^4+2 $ of length $l=10$, see Can you solve the listed smallest open Diophantine equations?. These equations have degree greater than $3$, which motivates studying the same problem for cubic equations.

After this previous equation has been solved, the equation (1) is the only cubic equation of length $l\leq 12$ for which it is open whether it is solvable in integers.

My attempt for solving (1) is: first observe that $x$ is odd. Then multiply the equation by $x$ and write $t=xy$ to get $$ x+4x^4+t^2+2tz^2=0, \quad\quad\quad\quad (2) $$ for which we are looking for solutions such that $x$ is odd and is a divisor of $t$. Interestingly, my search so far returned no solutions to (2) with odd $x$ at all, even if we drop the divisibility condition. Obviously, if (2) indeed has no solutions with odd $x$, then we are done.

If you like, (2) can also be rewritten as $$ x+4x^4+Y^2-z^4=0 $$ where $Y=t+z^2$. However, I am not sure how to proceed from there.

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    $\begingroup$ $x + Y^2 = (z^2 - 2x^2)(z^2 + 2x^2)$. -x is a quadratic residue modulo $z^2 - 2x^2$ and $z^2 + 2x^2$. There are four cases depending on whether $y^2 - x > 0$ and whether $z$ is odd or even. I’m not sure if it’s possible to get a contradiction in all cases, but some cases are excluded by calculating Jacobi symbols. $\endgroup$
    – Denis Shatrov
    Commented Dec 8, 2023 at 20:42
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    $\begingroup$ In fact, this project is a fascinating journey of continuous learning! I know some methods, use them for solving equations in order, and find the simplest-looking equations for which these methods do not work. I then post these equations, and, if they are solved, I usually learn new method, apply it to the next equations, and then the whole process is iterated. As a result, I now have a large collection of methods for solving equations, and, for each method, a set of simplest-looking equations for which this method works but easier ones not. And a set of simplest-looking open equations. $\endgroup$
    – Bogdan Grechuk
    Commented Dec 9, 2023 at 11:56
  • $\begingroup$ OK, that's a fair answer. I apologize for my grumpy question. I guess solving individual diophantine equations isn't much to my taste, but that's a personal judgement. :) $\endgroup$
    – Satan's Minion
    Commented Dec 9, 2023 at 12:35
  • $\begingroup$ The solution of $2y^3+xy+x^4+1=0$. Note that $x$ is odd and therefore $y$ is even. Considering the equation modulo 4, one obtains $4\mid(xy+2)$, so $y\equiv_42$. It follows from $y\mid(x^4+1)$ that $y=2p_1\cdot\ldots\cdot p_k$, for some necessarily distinct primes$p_i\equiv_81$, and thus $y\equiv_{16}2$. Modulo 8, we get $8\mid(xy+2)$. So, $x\equiv_43$. But $t\mid(x^4+1)$, where $0\ne t=x+2y^2\equiv_43$, so contradiction. $\endgroup$
    – te4
    Commented Dec 11, 2023 at 20:28
  • $\begingroup$ te4: Note that $yt=-(x^4+1)<0$, hence $y$ and $t$ have opposite sign. If $y>0>t$ then $t$ being $3$ mod $4$ is not a contradiction. If $t>0>y$, then $|y|$ is $2$ modulo $16$, and $y$ is $14$ modulo $16$, in which case $x$ and $t$ are $1$ modulo $8$, again no contradiction. $\endgroup$
    – Bogdan Grechuk
    Commented Dec 12, 2023 at 11:04

1 Answer 1

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Jacobi symbols help again. Let $x' = -x$, $z = 2z'$. In each case, we have enough information modulo 4 or 8 to calculate Jacobi symbol.

case 1. $x > 0$, $z \equiv 1 \pmod{2}$ $$ x + Y^2 = (z^2 - 2x^2)(z^2 + 2x^2)$$ $$\left(\frac{-x}{z^2 - 2x^2}\right) = -\left(\frac{z^2 - 2x^2}{x}\right) = -1$$

case 2. $-Y^2 < x < 0$, $z \equiv 1 \pmod{2}$ $$ Y^2 - x' = (z^2 - 2x'^2)(z^2 + 2x'^2)$$ $$\left(\frac{x'}{z^2 - 2x'^2}\right) = -\left(\frac{z^2 - 2x'^2}{x'}\right) = -1$$

case 3. $x < -Y^2$, $z \equiv 1 \pmod{2}$ $$ x' - Y^2 = (2x'^2 - z^2)(2x'^2 + z^2)$$ $$\left(\frac{x'}{2x'^2 - z^2}\right) = \left(\frac{2x'^2 - z^2}{x'}\right) = -1$$

case 4. $x > 0$, $z \equiv 0 \pmod{2}$ $$x + Y^2 = 4(2z'^2 - x^2)(2z'^2 + x^2)$$ $$\left(\frac{-x}{2z'^2 - x^2}\right) = \left(\frac{2z'^2 - x^2}{x}\right) = \left(\frac{2}{x}\right) = -1$$

case 5. $-Y^2 < x < 0$, $z \equiv 0 \pmod{2}$ $$Y^2 - x' = 4(2z'^2 - x'^2)(2z'^2 + x'^2)$$ $$\left(\frac{x'}{2z'^2 - x'^2}\right) = \left(\frac{2z'^2 - x'^2}{x'}\right) = \left(\frac{2}{x'}\right) = -1$$

case 6. $x < -Y^2$, $z \equiv 0 \pmod{2}$ $$x' - Y^2 = 4(x'^2 - 2z'^2)(x'^2 + 2z'^2)$$ $$\left(\frac{x'}{x'^2 - 2z'^2}\right) = \left(\frac{x'^2 - 2z'^2}{x'}\right) = \left(\frac{-2}{x'}\right) = -1$$

I suspect there is more elegant solution, but at least the equation is solved now.

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