Diophantine equation 22

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$${Published}$$  $${Online}$$  $${First}$$  $${(11/2/2024)}$$
$${Latest}$$  $${additions}$$  $${(11/2/2024)}$$
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$${Diophantine}$$ $${equation}$$ $${22}$$
$${(22.1)}$$  $${~~~\sum\limits_{i=1}^5\ a_i^3}$$ $${=}$$ $${n}$$ $${~~~(n  \in  \mathbb{N})}$$
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$${case(22.1)}$$  $${~~~\sum\limits_{i=1}^5\ a_i^3}$$ $${=}$$ $${n}$$ $${~~~(n  \in  \mathbb{N})}$$
次の恒等式から始める。
$${6r=(r+1)^3+(r-1)^3-2r^3}$$
ここで両辺に$${(-6t+j)^3}$$を足すと左辺は
$${6(r-36t^3+18jt^2-3j^2t)+j^3}$$となる。
ここで$${~~j^3 \equiv j \pmod 6}$$になるので、
$${(r-36t^3+18jt^2-3j^2t)=m_j}$$とすると
$${6m+j^3 \equiv 6m_j+ j \pmod 6}$$となる。
$${r=36t^3-18jt^2+3j^2t+m_j}$$

$${\begin{aligned}6m+j^3&=(36t^3-18jt^2+3j^2t+m_j+1)^3\\&+(36t^3-18jt^2+3j^2t+m_j-1)^3\\&+2(-36t^3+18jt^2-3j^2t-m_j)^3\\&+ (-6t+j)^3\end{aligned}}$$

$${j=0,1,2,3,-2(4),-1(5)}$$
を代入していくと解になる。
$${j=0→6m+j^3=6m_0}$$
$${j=1→6m+j^3=6m_1+1}$$
$${\begin{aligned}j=2→6m+j^3&=6(m_2+1)+2\\m_2&=m-1\end{aligned}}$$
$${\begin{aligned}j=3→6m+j^3&=6(m_3+4)+3\\m_3&=m-4\end{aligned}}$$
$${\begin{aligned}j=-2→6m+j^3&=6(m_4-2)+4\\m_4&=m+2\end{aligned}}$$
$${\begin{aligned}j=-1→6m+j^3&=6(m_5-1)+5\\m_5&=m+1\end{aligned}}$$

$${\begin{aligned}6m&=(36t^3+m+1)^3+(36t^3+m-1)^3\\&+2(-36t^3-m)^3 + (-6t)^3\end{aligned}}$$

$${\begin{aligned}6m+1&=(36t^3-18t^2+3t+m+1)^3\\&+(36t^3-18t^2+3t+m-1)^3\\&+2(-36t^3+18t^2-3t-m)^3\\&+ (-6t+1)^3\end{aligned}}$$

$${\begin{aligned}6m+2&=(36t^3-36t^2+12t+m)^3\\&+(36t^3-36t^2+12t+m-2)^3\\&+2(-36t^3+36t^2-12t-m+1)^3\\&+ (-6t+2)^3\end{aligned}}$$

$${\begin{aligned}6m+3&=(36t^3-54t^2+27t+m-3)^3\\&+(36t^3-54t^2+27t+m-5)^3\\&+2(-36t^3+54t^2-27t-m+4)^3\\&+ (-6t+3)^3\end{aligned}}$$

$${\begin{aligned}6m+4&=(36t^3+36t^2+12t+m+3)^3\\&+(36t^3+36t^2+12t+m+1)^3\\&+2(-36t^3-36t^2-12t-m-2)^3\\&+ (-6t-2)^3\end{aligned}}$$

$${\begin{aligned}6m+5&=(36t^3+18t^2+3t+m+2)^3\\&+(36t^3+18t^2+3t+m)^3\\&+2(-36t^3-18t^2-3t-m-1)^3\\&+ (-6t-1)^3\end{aligned}}$$

以上により、全ての自然数$${n}$$が$${5}$$個の立方数の和で表される事が示せた。
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因みに以下の解も知られています。
$${6r=(r+1)^3+(r-1)^3-2r^3}$$
$${6r±1=(r+1)^3 + (r−1)^3−2r^3 ± 1}$$
$${6r±2=r ^3+(r ± 2)^3−2(r ± 1)^3 \mp 2^3}$$
$${\begin{aligned}6r+3&=(r−3)^3+(r−5)^3\\&−2(r−4)^3+3^3\end{aligned}}$$
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$${REFERENCES}$$
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$${Ajai~Choudhry}$$
$${「Expressing~an~integer~as~a~\\sum~of~cubes~of~polynomials」}$$
$${arXiv:2311.07325~v1~[math.NT]~\\on~13~Nov ~2023}$$
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$${~~~\sum\limits_{i=1}^3\ a_i^3}$$ $${=}$$ $${n}$$ $${~~~(n  \in  \mathbb{N})→}$$こちらへ
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$${~~~\sum\limits_{i=1}^4\ a_i^3}$$ $${=}$$ $${n}$$ $${~~~(n  \in  \mathbb{N})→}$$こちらへ
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【目次001】・【目次002】・【目次003】
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