# All integer is a sum of four cubes of rational

I am interested in a classic problem of representation of integers as a sum of four cubes of integers. This problem has been partially solved missing only integers of the form 9k ± 4. I got a proof that this is true for rational. This is not important really because it is not given in a ring but in a field (where the operability is easier). From this the following question:

Prove that all rational integer n is a sum of four cubes of nonzero rational numbers.

• What is a rational integer? – Wojowu Apr 21 '15 at 15:56
• Any element of Z, not only positive integers. Actually one can ask just for natural numbers because -1 is a cube. – Piquito Apr 21 '15 at 17:25
• @Wojowu A rational integer is an integer in $\mathbb{Q}$, in contrast to integers from other "rings of integers" $\mathcal{O}_K$ of a number field $K$. – Dietrich Burde Apr 21 '15 at 18:19

Theorem (Reyley 1825): Every rational number $r$ can be represented as the sum of three rational cubes: $$r=x^3+y^3+z^3.$$ A short proof can be found in the artcle of Richmond (1930).
For $a\in \mathbb{Q}$ we can represent $s-a^3=x^3+y^3+z^3$, so that $s=a^3+x^3+y^3+z^3$ for any rational $s$.
• Your substitution s - $a^3$ shows -unintentionally I guess- in a very elementary way what is inferred at first level of elliptic curves theory: the representations for each number n here are infinitely many (which also applies to three cubes of Ryley of course). – Piquito Apr 21 '15 at 20:25