In Honour of Helmut Hasse.

It is well known that if $n \equiv \pm 4 \bmod 9$ , it cannot be represented as the sum of 3 cubes.

Basically, if $n=9k+4$ or $n=9k+5$ , where $k$ is an integer, $n$ cannot be represented as the sum of 3 cubes.

Since all the numbers in the list can be represented in the form $9k+4$ or $9k+5$ , all of them cannot be represented as the sum of 3 cubes.

Proof that all integers in the form of $9k+4$ or $9k+5$ cannot be represented as the sum of 3 cubes.

Consider the remainders of $x^3$ , $y^3$ and $z^3$ when divided by 9.

$\begin{array} { r l } x^3 \equiv (1,8,0) \bmod 9 \\ y^3 \equiv (1,8,0) \bmod 9 \\ z^3 \equiv (1,8,0) \bmod 9 \\ \end{array}$

Now, choose one remainder from $x^3$ , $y^3$ and $z^3$ and add them together.

For example, someone could choose the remainder of 1 from $x^3$ , the remainder of 8 from $y^3$ and the remainder of 8 from $z^3$ .

Adding the remainders together would get $1+8+8=17$ , which is equal to $8 \bmod 9$ .

Now, consider all the possible sums of remainders that you can get.

It turns out that ${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }\equiv \quad 0,1,2,3,6,7,8\quad mod\quad 9$ .

It looks like that $4$ and $5$ are not in the list, indicating that the sum of 3 cubes can never give a remainder of $4$ or $5$ , thus completing the proof.