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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

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$^*$ 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

$3^6(a^2+b^2+c^3+d^3)=(27a)^2+(27b)^2+(9c)^3+(9d)^3$.

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You do not need the final $d^3.$ Every integer is the sum of two squares and a cube, as long as we do not restrict the $\pm$ sign on the cube.

TYPESET FOR LEGIBILITY:

Solution by Andrew Adler:

$$ 2x+1 = (x^3 – 3 x^2 + x)^2 +(x^2 – x – 1)^2 -(x^2 – 2x)^3 $$

$$ 4x+2 = (2x^3 – 2 x^2 – x)^2 +(2x^3 -4x^2 – x + 1)^2 -(2x^2 – 2x-1)^3 $$

$$ 8x+4 = (x^3 + x +2 )^2 +(x^2 – 2x – 1)^2 -(x^2 + 1)^3 $$

$$ 16x+8 = (2x^3 – 8 x^2 +4 x +2)^2 +(2x^3 -4x^2 – 2 )^2 -(2x^2 – 4x)^3 $$

$$ 16x = (x^3 +7 x – 2)^2 +(x^2 +2 x + 11)^2 -(x^2 +5)^3 $$

You can check these with your own computer algebra system. Please let me know if I mistyped anything.

Alright, our conjecture (Kaplansky and I) is that, for any odd prime $q,$ $x^2 + y^2 + z^q$ is universal. However, this is false as soon as the exponent on $z$ is odd but composite. The example we put in the article is

$$ x^2 + y^2 + z^9 \neq 216 p^3, $$

where $p \equiv 1 \pmod 4$ is a (positive) prime.

This defeated a well-known conjecture of Vaughan. We told him about it in time for him to include it in the **second edition** of his HARDY-LITTLEWOOD BOOK, where it is now mentioned on pages 127 (“There are some exceptions to this,”) and exercise 5 on page 146.

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