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Prove that there exist infinitely many integers $x$ such that integer $P(x)=x^6-4$ is a sum of two squares of integers.

Equivalently, prove that $x^3-2$ and $x^3+2$ are simultaneously sums of two squares for infinitely many integers $x$.

This is an extension of my previous question about $P(x)=x^3-2$.

The methods in the answers to that question are sufficient to resolve this problem for all examples of cubic and quartic polynomials $P(x)$ with small coefficients (although I do not have any general theorem that they always work). However, I do not see how to apply such methods for higher degree polynomials, and $P(x)=x^6-4$ is one of the nicest examples.

By Schinzel's hypothesis H, $x^3-2$ and $x^3+2$ are simultaneously primes infinitely often. If $x$ is $3$ modulo $4$, these primes are both $1$ modulo $4$, and the claim follows.

But Schinzel's hypothesis H is out of reach. There are many papers proving that values of certain polynomials are the products of a bounded number of primes, but I never saw results of this type when the primes are restricted to be $1$ modulo $4$.

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This problem looks tricky. I'd recommend you look up "Châtelet surfaces"; these are a slightly easier case when one has a polynomial of degree $4$ instead of a polynomial of degree $6$, but a lot of the geometry is similar (both are conic bundle surfaces).

A slightly easier question would be: Are there infinitely many rational numbers $x$ such that $x^6 - 4$ is a sum of two rational squares? The answer to this is yes, but I'm not sure of an elementary proof of this. It follows from the fact that the corresponding conic bundle surfaces are unirational; this is a special case of a result of Kollár and Mella [1, Corollary 8] (your conic bundle has $6$ singular fibres).

Your specific equation over the rational numbers is also considered in the paper [2] by Swinnerton-Dyer. Possibly the method of Swinnerton-Dyer could be adapted to handle your problem involving integer $x$.

I'd be very happy to see an elementary answer which avoids all this geometry!

[1] Kollár, János; Mella, Massimiliano. Quadratic families of elliptic curves and unirationality of degree 1 conic bundles. Amer. J. Math. 139 (2017), no. 4, 915--936.

[2] Swinnerton-Dyer, Peter. Rational points on some pencils of conics with 6 singular fibres. Ann. Fac. Sci. Toulouse Math. (6) 8 (1999), no. 2, 331--341.

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    $\begingroup$ If $n^6 - 4 = a^2 + b^2$, then $(n^4/2)^6 - 4 = \frac{1}{64}(n^{12} + 16)(n^6 + 4)(n^6 - 4)$ is a sum of squares also. But this resolves the problem only for rational numbers, because $x^6 - 4$ can not be a sum of squares for even integer $x$. $\endgroup$ Commented Sep 9, 2023 at 10:07

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