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Let $A=\{x^4+y^4:\ x,y\in\mathbb Q\}$. Then $$A-A:=\{a-b:\ a,b\in A\}=\{u^4+v^4-x^4-y^4:\ u,v,x,y\in\mathbb Q\}.$$

Motivated by Question 415482, here I ask the following question.

Question. Is it true that $A-A=\mathbb Q$? Any effective way to approach it？

By my computation, $A-A$ at least contains $0,1,\ldots,562$. For example, $$248=\left(\frac{95}{28}\right)^4+\left(\frac{135}{14}\right)^4-\left(\frac{13}7\right)^4-\left(\frac{269}{28}\right)^4\in A-A.$$ From the viewpoint of additive combinatorics, the question looks interesting. I guess that it should have a positive answer. Any ideas to solve it?

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According to Tito Piezas's website $x^4+y^4-(z^4+t^4) = N$, There is an identity $((2a+b)c^3d)^4 + (2ac^4-bd^4)^4 - (2ac^4+bd^4)^4 - ((2a-b)c^3d)^4 = a(2bcd)^4$ where $b = c^8-d^8$, for arbitrary {$a,c,d$}.

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    $\begingroup$ Tomita, thank you for pointing out the identity. Would you please proivde a website link for that identity? $\endgroup$
    – Zhi-Wei Sun
    Commented Feb 7, 2022 at 13:15
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    $\begingroup$ See oeis.org/A351306 for a stronger conjecture of mine: Each $n\in\mathbb N$ can be written as $u^4+v^4-(x^4+y^4)$ with $u,v,x,y\in\mathbb Q$ and $x^4+y^4\le n^2$. Can this be explained via the identity? $\endgroup$
    – Zhi-Wei Sun
    Commented Feb 7, 2022 at 16:09
  • $\begingroup$ The equation is the link. Scroll down to #4. :) $\endgroup$
    – Wolfgang
    Commented Feb 7, 2022 at 21:32
  • $\begingroup$ @Tomita Given link is not working $\endgroup$
    – Guruprasad
    Commented Dec 5 at 0:48
  • $\begingroup$ @Guruprasad, I edited it. $\endgroup$
    – Tomita
    Commented Dec 5 at 2:23

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