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It's easy$^*$ to prove that if $n=3^{6m}(3k \pm 1)$ where $(m,k) \in \mathbb{N} \times \mathbb{Z}$, then $n=a^2+b^2+c^3+d^3$ with $(a,b,c,d) \in \mathbb{Z}^4$. But how to prove that this is true if $n=3k$?

Thanks,

W

$^*$ Because $3k+1=0^2+(3k+8)^2+(k+1)^3+(-k-4)^3$, $3k+2=1^2+(3k+8)^2+(k+1)^3+(-k-4)^3$ and $3^6(a^2+b^2+c^3+d^3)=(27a)^2+(27b)^2+(9c)^3+(9d)^3$.

  • 0
    What is $\mathbb Z^4$?2012-07-23
  • 4
    @ChuckFernández I presume the set of all 4-tuples comprised of integers?2012-07-23
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    does it mean the same as $a,b,c,d\in \mathbb Z$?2012-07-23
  • 0
    @ChuckFernández Yes.2012-07-23

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