丢番图方程组的正整数本原解
本帖最后由 葡萄糖 于 2020-3-31 12:23 编辑已知正整数 \(a\), \(b\), \(c\), \(p\), \(q\) 满足如下丢番图方程组
\begin{align*}
\Large
\left\{
\begin{split}
a^2+b^2+c^2=p^2\\
a^3+b^3+c^3=q^3
\end{split}
\right.
\end{align*}
其中本原解 \(\big(a,\,b,\,c\big)\) 满足 \(\gcd\{a,\,b,\,c\}=1\).
请问在 \(\max\{a,b,c\}<1000000\) 的前提下，该丢番图方程组是否只有16组本原解？
① {14, 23, 70},
② {3, 34, 114},
③ {18, 349, 426},
④ {145, 198, 714},
⑤ {213, 3450, 3466},
⑥ {1016, 2364, 3693},
⑦ {3542, 9286, 13009},
⑧ {29233, 29574, 44754},
⑨ {6213, 32194, 46458},
⑩ {32913, 38444, 54264},
⑪ {106677, 117252, 124876},
⑫ {110051, 118044, 295512},
⑬ {116457, 286752, 436136},
⑭ {202023, 234550, 510270},
⑮ {43472, 613560, 628485},
⑯ {272267, 417416, 875656}
A^2 + B^2 + C^2 = Square, A^3 + B^3 + C^3 = Cube. A Martin, 1898
http://www.mathpuzzle.com/cbumpkin.txt
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