x^3+y^4=z^5 的正整数解

王守恩

这题目有点沉，不是有点沉，而是沉得很：

1, 对一组数，我们还找不到是否有解(譬如:2, 5, 10)。

2, 对有解的一组数，我们还找不到最小解。

希望还是在，对特殊的一组数, 我们还是能找到最小解：

\(x^a+y^b=z^c\)  满足:  \(u=v*c-1=s*LCM(a,b)\)

则: \(x=2^{u/a},y=2^{u/b},z=2^{v}\) 是最小解。

谢谢 gxqcn！谢谢3楼的第3个公式。

王守恩

有这样一串数:  \(a(n)=\)\(\big\lfloor\frac{n^2}{4}\big\rfloor\)     n=1, 2, 3, 4, 5, 6, ......   

0, 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90, 100, 110, 121, 132, ......

求  \(x^{a(n)}+y^{a(n+1)}=z^{a(n+2)}\)  正整数解。

\(a(n),a(n+1),a(n+2)\)  可以交换位置。

王守恩

有这样一串数:  \(a(n)=\)\(\frac{n(n-1)}{2}\)     n=1, 2, 3, 4, 5, 6, ......

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, ......  

求  \(x^{a(n)}+y^{a(n+1)}=z^{a(n+2)}\)  正整数解。

\(a(n),a(n+1),a(n+2)\)  可以交换位置。

王守恩

求证:\(\D\frac{\big(2^{(3*2^{2n}+2^{n+2}+1)k+3*2^{2n-1}+2^{n-1}}\big)^{2^{n+1}+1}+\big(2^{(2^{2n+1}+3*2^{n}+1)k+2^{2n}+2^{n-1}}\big)^{3*2^{n}+1}}{\big(2^{(3*2^{2n+1}+5*2^{n}+1)k+3*2^{2n}-2^{n-1}+1}\big)^{2^{n}+1}}=1\)

n=0, 1, 2, 3, 4, 5, 6...        k=0, 1, 2, 3, 4, 5, 6, ...

王守恩

求\(x^3+y^4=z^5\) 的正整数解，基本也就这两种解法。

1,\(\frac{\big((v^{5 a} - u^{4 b})^{7} k^{20 n}\big)^{3} +\big (u^{ b} (v^{5 a} - u^{4 b})^{5} k^{15 n}\big)^{4}}{\big(v^{a} (v^{5 a} - u^{4 b})^{4} k^{12 n}\big)^{5}}=1\)
   v=2, 3, 4, 5, 6, 7, 8, 9, ......
   u=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   a=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   b=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   n=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   k=1, 2, 3, 4, 5, 6, 7, 8, 9, ......


2,\(\frac{\big(v^a (v^{3 a} + u^{4 b})^{8} k^{20 n}\big)^{3} +\big (u^ {b} (v^{3 a} + u^{4 b})^{6} k^{15 n}\big)^{4}}{\big((v^{3 a} + u^{4 b})^{5} k^{12 n}\big)^{5}}=1\)
   v=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   u=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   a=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   b=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   n=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   k=1, 2, 3, 4, 5, 6, 7, 8, 9, ......

王守恩

\(\frac{\big (u^{ b} (v^{(2n+3)a} - u^{(2n+1)b})^{2n+3}\big)^{2n+1}+\big((v^{(2n+3) a} - u^{(2n+1) b})^{2n+2}\big)^{2n+2}}{\big(v^{a} (v^{(2n+3)a} - u^{(2n+1) b})^{2n+1}\big)^{2n+3}}=1\)
   v=2, 3, 4, 5, 6, 7, 8, 9, ......
   u=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   a=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   b=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   n=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......

\(\frac{\big (u^{ b} (v^{(s*n+t+1)a} - u^{(s*n+t-1)b})^{s*n+t+1}\big)^{s*n+t-1}+\big((v^{(s*n+t+1) a} - u^{(s*n+t-1) b})^{s*n+t}\big)^{s*n+t}}{\big(v^{a} (v^{(s*n+t+1)a} - u^{(s*n+t-1) b})^{s*n+t-1}\big)^{s*n+t+1}}=1\)
   v=2, 3, 4, 5, 6, 7, 8, 9, ......
   u=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   a=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   b=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   n=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   s=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   t=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......

王守恩

\(\frac{\big (u^{ b} (v^{(s*n+t+1)a} - u^{(s*n+t-1)b})^{s*n+t+1}*k^{(s n + t) (s n + t + 1) p}\big)^{s*n+t-1}+\big((v^{(s*n+t+1) a} - u^{(s*n+t-1) b})^{s*n+t}*k^{(s n + t - 1) (s n + t + 1) p}\big)^{s*n+t}}{\big(v^{a} (v^{(s*n+t+1)a} - u^{(s*n+t-1) b})^{s*n+t-1}*k^{(s n + t - 1) (s n + t) p}\big)^{s*n+t+1}}=1\)
   v=2, 3, 4, 5, 6, 7, 8, 9, ......
   u=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   a=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   b=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   n=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   s=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   t=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   k=1, 2, 3, 4, 5, 6, 7, 8, 9, ......
   p=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ......

王守恩

\(\big((a^{2}-1)^{08}\big)^{1}+\big((a^{2}-1)^{03}\big)^{03}=\big(a(a^{2}-1)^{04}\big)^{2}\)
\(\big((a^{3}-1)^{12}\big)^{2}+\big((a^{3}-1)^{05}\big)^{05}=\big(a(a^{3}-1)^{08}\big)^{3}\)
\(\big((a^{4}-1)^{16}\big)^{3}+\big((a^{4}-1)^{07}\big)^{07}=\big(a(a^{4}-1)^{12}\big)^{4}\)
\(\big((a^{5}-1)^{20}\big)^{4}+\big((a^{5}-1)^{09}\big)^{09}=\big(a(a^{5}-1)^{16}\big)^{5}\)
\(\big((a^{6}-1)^{24}\big)^{5}+\big((a^{6}-1)^{11}\big)^{11}=\big(a(a^{6}-1)^{20}\big)^{6}\)
\(\big((a^{7}-1)^{28}\big)^{6}+\big((a^{7}-1)^{13}\big)^{13}=\big(a(a^{7}-1)^{24}\big)^{7}\)
\(\big((a^{8}-1)^{32}\big)^{7}+\big((a^{8}-1)^{15}\big)^{15}=\big(a(a^{8}-1)^{28}\big)^{8}\)
\(\big((a^{9}-1)^{36}\big)^{8}+\big((a^{9}-1)^{17}\big)^{17}=\big(a(a^{9}-1)^{32}\big)^{9}\)

王守恩

\(\big((a^{2}-5)^{08}\big)^{1}+\big((a^{2}-5)^{04}\big)^{02}+\big((a^{2}-5)^{03}\big)^{03}+\big((a^{2}-5)^{02}\big)^{04}+\big((a^{2}-5)^{2}\big)^{04}+\big((a^{2}-5)^{1}\big)^{08}=\big(a(a^{2}-5)^{04}\big)^{2}\)
\(\big((a^{3}-5)^{12}\big)^{2}+\big((a^{3}-5)^{06}\big)^{04}+\big((a^{3}-5)^{05}\big)^{05}+\big((a^{3}-5)^{04}\big)^{06}+\big((a^{3}-5)^{3}\big)^{08}+\big((a^{3}-5)^{2}\big)^{12}=\big(a(a^{3}-5)^{08}\big)^{3}\)
\(\big((a^{4}-5)^{16}\big)^{3}+\big((a^{4}-5)^{08}\big)^{06}+\big((a^{4}-5)^{07}\big)^{07}+\big((a^{4}-5)^{06}\big)^{08}+\big((a^{4}-5)^{4}\big)^{12}+\big((a^{4}-5)^{3}\big)^{16}=\big(a(a^{4}-5)^{12}\big)^{4}\)
\(\big((a^{5}-5)^{20}\big)^{4}+\big((a^{5}-5)^{10}\big)^{08}+\big((a^{5}-5)^{09}\big)^{09}+\big((a^{5}-5)^{08}\big)^{10}+\big((a^{5}-5)^{5}\big)^{16}+\big((a^{5}-5)^{4}\big)^{20}=\big(a(a^{5}-5)^{16}\big)^{5}\)
\(\big((a^{6}-5)^{24}\big)^{5}+\big((a^{6}-5)^{12}\big)^{10}+\big((a^{6}-5)^{11}\big)^{11}+\big((a^{6}-5)^{10}\big)^{12}+\big((a^{6}-5)^{6}\big)^{20}+\big((a^{6}-5)^{5}\big)^{24}=\big(a(a^{6}-5)^{20}\big)^{6}\)
\(\big((a^{7}-5)^{28}\big)^{6}+\big((a^{7}-5)^{14}\big)^{12}+\big((a^{7}-5)^{13}\big)^{13}+\big((a^{7}-5)^{12}\big)^{14}+\big((a^{7}-5)^{7}\big)^{24}+\big((a^{7}-5)^{6}\big)^{28}=\big(a(a^{7}-5)^{24}\big)^{7}\)
\(\big((a^{8}-5)^{32}\big)^{7}+\big((a^{8}-5)^{16}\big)^{14}+\big((a^{8}-5)^{15}\big)^{15}+\big((a^{8}-5)^{14}\big)^{16}+\big((a^{8}-5)^{8}\big)^{28}+\big((a^{8}-5)^{7}\big)^{32}=\big(a(a^{8}-5)^{28}\big)^{8}\)
\(\big((a^{9}-5)^{36}\big)^{8}+\big((a^{9}-5)^{18}\big)^{16}+\big((a^{9}-5)^{17}\big)^{17}+\big((a^{9}-5)^{16}\big)^{18}+\big((a^{9}-5)^{9}\big)^{32}+\big((a^{9}-5)^{8}\big)^{36}=\big(a(a^{9}-5)^{32}\big)^{9}\)

王守恩

加数指数是连续的，还有吗？

\(\big(4^{024}\big)^{1}+\big(4^{012}\big)^{2}+\big(4^{008}\big)^{03}+\big(4^{006}\big)^{04}=\big(4^{05}\big)^{05}\)

\(\big(4^{060}\big)^{2}+\big(4^{040}\big)^{3}+\big(4^{030}\big)^{04}+\big(4^{024}\big)^{05}=\big(4^{11}\big)^{11}\)

\(\big(4^{120}\big)^{3}+\big(4^{090}\big)^{4}+\big(4^{072}\big)^{05}+\big(4^{060}\big)^{06}=\big(4^{19}\big)^{19}\)

\(\big(4^{210}\big)^{4}+\big(4^{168}\big)^{5}+\big(4^{140}\big)^{06}+\big(4^{120}\big)^{07}=\big(4^{29}\big)^{29}\)

\(\big(4^{336}\big)^{5}+\big(4^{280}\big)^{6}+\big(4^{240}\big)^{07}+\big(4^{210}\big)^{08}=\big(4^{41}\big)^{41}\)

\(\big(4^{504}\big)^{6}+\big(4^{432}\big)^{7}+\big(4^{378}\big)^{08}+\big(4^{336}\big)^{09}=\big(4^{55}\big)^{55}\)

\(\big(4^{720}\big)^{7}+\big(4^{630}\big)^{8}+\big(4^{560}\big)^{09}+\big(4^{504}\big)^{10}=\big(4^{71}\big)^{71}\)

\(\big(4^{990}\big)^{8}+\big(4^{880}\big)^{9}+\big(4^{792}\big)^{10}+\big(4^{720}\big)^{11}=\big(4^{89}\big)^{89}\)