- $2n+1 = (n^2-n-1)^2+(n^3-3n^2+n)^2+(2n-n^2)^3$ - $4n+2 = (2n^3-2n^2-n)^2+(2n^3-4n^2-n+1)^2+(-2n^2+2n+1)^3$ - $8n+4 = (n^2-2n-1)^2+(n^3+n+2)^2+(-n^2-1)^3$ - $8(a^2+b^2+c^3) = (2a+2b)^2+(2a-2b)^2+(2c)^3$ Thus we can write every integer $n$ as $n=8^k \cdot s$, $s \not\equiv 0 \mod 8$, then use the first three identities to represent $s$ and then the last one $k$ times to represent $n$.