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Hasse principle is known to hold for homogeneous quadratic equations, but fail for some 3- and 4-variable cubics, such as $5x^3+4y^3+3z^3=0$ or $15x^3+10y^3+4z^3+3t^3=0$. These counterexamples are explained by Brauer–Manin obstructions.

It is known that that homogeneous equations in $n\geq 5$ variables of degree $d<n$ defining smooth surfaces do not have Brauer–Manin obstructions [1]. Hence, any such counterexample to the Hasse principle would be very interesting (and would contradict conjectures stating that Brauer–Manin obstructions are the only ones for certain families of equations).

For example, consider family of equations $ax^4+by^4+cz^4+dt^4+es^4=0$ in $n=5$ variables of degree $d=4$. If we order these equations by $|abcde|$, then the first interesting one is the equation $$ x^4 + y^4 + z^4 + t^4 = 36 s^4, $$ which leads to the question in the title. If we order these equations by $|a|+|b|+|c|+|d|+|e|$, then the first interesting one is $$ 7 x^4 + 3 y^4 + 3 z^4 + 3 t^4 = 2 s^4 $$ which is equivalent to $$ 189 x^4 + y^4 + z^4 + t^4 = 54 s^4. $$ Here, by "interesting" I mean that these equations are locally solvable modulo small primes but have no small integer solutions.

My specific question is whether these equations have any integer solutions $(x,y,z,t,s)\neq (0,0,0,0,0)$. My general question is whether there are any techniques for searching large solutions to equations like this better than direct search combined with analysis modulo small integers? Here, by "direct search" I mean trying values of 4 variables and solve the resulting equations in 1 variable. If there are better methods, are they implemented in any computer algebra system, e.g. Magma?

[1] Bjorn Poonen and Jos´e Felipe Voloch. Random Diophantine equations. In Arithmetic of higher-dimensional algebraic varieties (Palo Alto, CA, 2002), volume 226 of Progr. Math., pages 175–184. Birkh¨auser Boston, Boston, MA, 2004. With appendices by Jean-Louis Colliot-Th´el`ene and Nicholas M. Katz.

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    $\begingroup$ Do you already know that $x^4+y^4+z^4+t^4=36$ admits a $p$-adic solution for all p? You write "modulo small primes", but not "modulo all primes". (At $p=2$ it suffices to find a solution mod 16, which is easy to do, and at all other primes we are looking for a solution mod $p$) $\endgroup$
    – Achim Krause
    Commented May 29 at 15:28
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    $\begingroup$ No, I do not know how to check the local solubility for all $p$ at once, so I just checked it for small primes. $\endgroup$
    – Bogdan Grechuk
    Commented May 29 at 15:48
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    $\begingroup$ Ah, here's an argument: Let $P\subseteq \mathbb{F}_p$ be the subset of $4$-th powers. Then $|P|=(p-1)/4 +1$ or $|P|=(p-1)/2+1$ depending on whether $p=1$ or $3$ mod $4$. By the Cauchy-Davenport inequality, it follows that $P+P+P+P$ contains all $p$ elements. So the equation has a solution mod every odd prime (and with a bit of extra manual work at $p=2$ and $p=3$, a solution in $\mathbb{Z}_p$ for any $p$.) $\endgroup$
    – Achim Krause
    Commented May 29 at 18:10
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    $\begingroup$ @AchimKrause Wouldn't Chevalley-Warning also apply in this case? $\endgroup$
    – R.P.
    Commented May 29 at 19:50
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    $\begingroup$ Yes. Specifically, if you have a solution $x^4 + y^4 + z^4 + t^4 = 4$ mod $16$, then one of $x,y,z,t$ needs to be odd, say $x$. Now by a version of Hensel's lemma, it is possible to deform $x$ in $\mathbb{Z}_2$ to an $\widetilde{x}=x + 4h$ with $\widetilde{x}^4 + y^4 + z^4 + t^4 = 36$ in $\mathbb{Z}_2$. $\endgroup$
    – Achim Krause
    Commented May 30 at 9:03

1 Answer 1

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The answer is "yes": $$ \left(\frac{1315}{1489}\right)^4 + \left(\frac{1843}{1489}\right)^4 + \left(\frac{2195}{1489}\right)^4 + \left(\frac{3435}{1489}\right)^4 = 36. $$

The "general" question was whether there are any techniques for searching large solutions to equations like this better than trying values of 4 variables and solve the resulting equations in 1 variable.

The answer to this is also yes. The technique is very simple: select some bound $M$ and create a large list $L$ of values of $x^4+y^4$ with $0\leq x < M$ and $0\leq y < M$. Then try values of $z,t,s$ up to $M$, compute $36s^4-t^4-s^4$, and check whether it belongs to $L$.

For this method to work, it is crucial to check membership in a huge list very quickly. A fast way to do this in Mathematica is suggested at https://mathematica.stackexchange.com/questions/41753/fastest-way-to-check-for-list-membership We need to do a one-time operation

assoc = <|Thread[L -> True]|>;

and then membership of i can be very quickly tested using

Lookup[assoc, i, False]

The list $L$ contains less than $M^2$ elements, and we need to test less than $M^3$ candidate members of the form $36s^4-t^4-s^4$. For large $M$, this is a huge gain in comparison to trying $M^4$ combinations of $x,y,z,t$ and solve for $s$.

The application of this method with $M=10000$ returned solution $$ x = 1315, y = 1843, z = 2195, t = 3435, s = 1489 $$ to the equation $$ x^4 + y^4 + z^4 + t^4 = 36 s^4 $$ after about 5-7 hours computation on standard PC.

The question also mentions equation $$ 7 x^4 + 3 y^4 + 3 z^4 + 3 t^4 = 2 s^4 $$ Using the same method, I also found a solution to this equation: $$ x = 2439, y = 1805, z = 2765, t = 4925, s = 5772. $$

This is not the first time I cannot solve a problem for a long time and then solve in a few days after asking on Mathoverflow.

If anyone knows even faster method for solution search for equations in $n\geq 4$ variables, please let me know in comments.

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