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Question: Do there exist integers $(x,y,z)\neq (0,0,0)$ such that $$ 13x^4+11y^4=8z^4 ? $$

Some motivation: This is currently the smallest (in a sense defined here On the smallest open Diophantine equations: beyond Hilbert's 10 problem) three-monomial homogeneous equation that I do not see how to solve. Search returns no non-trivial solutions. Magma says that the equation is locally solvable. Another standard method to solve equations of the form $ax^4+by^4+cz^4=0$ is to look at genus 1 curves $ax^2+by^4+cz^4=0$, $ax^4+by^2+cz^4=0$ and $ax^4+by^4+cz^2=0$, and if any one of them has rank $0$, the rest is easy. But in this example the ranks are $1,1,2$.

I found many papers studying equations $x^4+y^4=cz^4$ systematically for all $c$ up to $10,000$, but cannot find a similar systematic study of more general $ax^4+by^4+cz^4=0$.

Bogdan Grechuk
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    If I'm understanding things correctly, there should be only finitely many integral solutions to the first genus 1 curve you mentioned. There is an algorithm to find them all (assuming irreducibility over $\overline{\mathbb{Q}}$); see https://www.jstor.org/stable/4100027 although there may be easier algorithms for this specific curve. Then just check if any of those finitely many points has a square $x$ value. – Pace Nielsen Apr 21 '23 at 20:35
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    By the way, the title says "rational" but your actual question says "integer". – Pace Nielsen Apr 21 '23 at 20:37
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    The original equation is homogeneous, so existence of rational and integral solutions are equivalent. And these solutions maps to rational points on elliptic curves, and there are infinitely many such rational points. – Bogdan Grechuk Apr 21 '23 at 20:53

2 Answers2

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This question is amenable to the use of the Mordell-Weil sieve. (For a good introduction to this technique, see the paper here.) In this situation, there is a simple version of it (using a single prime) suffices.

Let $C : 13x^{4} + 11y^{4} = 8z^{4}$, and let $E : y^{2} = x^{3} + 3146x$. (This elliptic curve is isomorphic to $13x^{4} + 11y^{2} = 8$. I chose this elliptic curve because it has rank $1$.) Therefore, we have a map $\phi : C \to E$. For a prime $p$ for which $C$, $E$ and the map $\phi$ have good reduction, we obtain the following commutative diagram. $\require{AMScd}$ \begin{CD} C(\mathbb{Q}) @>\phi>> E(\mathbb{Q})\\ @VV r V @VV \beta V\\ C(\mathbb{F}_{p}) @>\alpha>> E(\mathbb{F}_{p}) \end{CD} The vertical maps are reduction modulo $p$, and the horizontal maps are the map from $C$ to $E$ (the top row in characteristic zero, the bottom row in characteristic $p$).

While we don't know $C(\mathbb{Q})$, the other three sets are computable. (In particular, $E(\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$ and is generated by $(0,0)$ and $(3041064/87025,10028940012/25672375)$.)

In the event that there is a rational point $Q$ in $C(\mathbb{Q})$, we have that $(\alpha \circ r)(Q) = (\beta \circ \phi)(Q)$ is in ${\rm im~} \alpha \cap {\rm im~} \beta$. So if ${\rm im~} \alpha \cap {\rm im~} \beta = \emptyset$, it follows that $C(\mathbb{Q})$ is empty. Computer calculations show that ${\rm im~} \alpha \cap {\rm im~} \beta = \emptyset$ for $p \in \{ 17, 197, 1321, 2089 \}$. It follows that $C(\mathbb{Q}) \subseteq \mathbb{P}^{2}(\mathbb{Q})$ is empty.

Even if a single prime didn't suffice, it would also be possible to combine congruence information from different primes to try to obtain a contradiction.

Peter Mueller
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Jeremy Rouse
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  • Thank you for the answer! Would you be able to share a code used for "computer calculations"? Are there standard functions (e.g. in Magma) for computing the image of, say, map $\beta$? – Bogdan Grechuk Apr 27 '23 at 16:00
  • Also, you write that vertical lines are reductions modulo $p$. If $E(\mathbb Q)$ contains some rational points such that their denominators (in irreducible form) are divisible by $p$, how such points should be reduced modulo $p$? For example, how can we reduce a fraction like $5/17$ modulo $17$? – Bogdan Grechuk Apr 27 '23 at 17:22
  • I've sent you the code that I used via email. (If anyone else wants it, feel free to contact me. It's probably a bit much to add to my answer.) Regarding your question about reduction mod $p$, the key is that we think of $E$ as a projective variety. Really $E$ is $y^{2} z = x^{3} + 3146xz^{2}$ and so any rational point on $E$ can be scaled so $x$, $y$ and $z$ are coprime integers. Then we can reduce these integers mod $p$. The consequence of this is that if $(x : y : 1)$ is a point on $E$ and $p$ divides the denominator of $x$ or $y$, the point reduces to $(0 : 1 : 0) \in E(\mathbb{F}_{p})$. – Jeremy Rouse Apr 27 '23 at 22:02
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Here it is a partial answer discarding cases and concluding that the equation has an integer solution only if $x$ and $z$ are odd numbers not divisible by $5$, and $y$ is some odd number divisible by $5$ (too long for a comment).

Suppose there exist integer solutions. As $8z^4$ is even, then $x$ and $y$ must be both odd or both even.

Consider that $x$, $y$ and $z$ are odd. It is not difficult to show that any odd number to the fourth power other than those ending in $5$ ends in $1$. Thus, if $x$, $y$ and $z$ were odd integers ending in an odd number other than $5$, the equation would not have a solution, as $13x^4$ would be some positive integer ending in $3$, $11y^4$ would be some positive integer ending in $1$, and $8z^4$ would be some positive integer ending in $8$. Following the same reasoning, if any of $x$, $y$ or $z$ were odd integers ending in an odd number other than $5$, and the others were odd integers ending in $5$, the equation would not have a solution. Thus, the only possibilities for $x$, $y$ and $z$ to be the three odd would be that they were odd numbers divisible by $5$, or that $y$ were divisible by $5$ and $x$ and $z$ not.

However, in the first case, we would have $$13(5x')^4+11(5y')^4=8(5z')^4$$ And dividing by $5^4$, $$13x'^4+11y'^4=8z'^4$$ Therefore, there must exist a primitive equation such that $x$, $y$ and $z$ are not the three odd numbers divisible by $5$.

Suppose that $z$ were even, and $x$ and $y$ were odd. In this case, $8z^4$ would be some number ending in $8$ no matter the value of $z$. Thus, the only possibility for the equation to hold would be $y$ to be some odd number divisible by $5$, and $x$ some odd number not divisible by $5$. However, if the equation holds, $13x^4+11y^4$ would be a number ending in 8 and divisible by 8, so $z^4$ would be some odd number ending in $1$, and thus $z$ some odd number other than $5$. As we stated that $z$ was even, we reach a contradiction.

Consider that both $x$ and $y$ are even. If $x$,$y$ and $z$ were all even, we could keep dividing each term of the equation by $2^4$ until we had a primitive equation $13x'^4+11y'^4=8z'^4$ were at least one of $x'$, $y'$ and $z'$ were odd.

Finally, consider that $z'$ were odd and both $x'$ and $y'$ were even. In this last case, dividing both sides of the equation by 8 we would get that the sum of two even numbers is equal to some odd number, reaching a final contradiction.

As there are no more possible cases left, it is proved that the equation $13x^4+11y^4=8z^4$ has integer solutions other than the trivial one $(0,0,0)$ only if $y$ is some odd number divisible by $5$, and $x$ and $z$ are odd numbers not divisible by $5$.

Juan Moreno
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