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, ...... |