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](https://mathoverflow.net/questions/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?



Does $A-A=\mathbb Q$ hold for $A=\{x^4+y^4:\ x,y\in\mathbb Q\}$?
According to Tito Piezas's website [$x^4+y^4-(z^4+t^4) = N$](https://sites.google.com/site/tpiezas/001b),
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 {$a,c,d$}.


