判定梅森质数的卢卡斯序列

Treenewbee

或者  \[\left(3^{15}\right)^5+\left(2*\ 3^{25}\right)^3=\left(3^{11}\right)^7\]

lusishun

wlc1 发表于 2024-2-4 23:48
求：\(a^3+b^5=c^7\)

三：
由a^84+b^85=c^84.得，，
x=（2^84-1)^28,
Y=(2^84-1)^17,
Z=[2(2^84-1)]^12.

(待验算）

lusishun

lusishun 发表于 2024-2-5 00:54
三：
由a^84+b^85=c^84.得，，
x=（2^84-1)^28,

Czl,您感觉不对，大概说是字母用混乱了的原因。答案中x,y,z,是原方程中的a,b,c

朱明君

\(\left( 2^{30}\right)^3+\left( 2^{18}\right)^5=\left( 2^{13}\right)^7\)

朱明君

\(\left( 2^7\right)^5+\left( 2^5\right)^7=\left( 2^4\right)^9\)

朱明君

\(\left( 2^7\right)^5+\left( 2^5\right)^7=\left( 2^6\right)^6\)

lusishun

(2^9)^11+(2^11)^9=(2^10)^10

lusishun

太多了，因为2024^2-1=4096575',
(2^2023)^2025+(2^2025)^2023=(2^2024)^2024.
(2^17)^240975+(2^240975)^17=(2^2024)^2024,
……………………………

朱明君

\(设n为正整数，\)
\(则\left( 2^{n+2}\right)^n+\left( 2^n\right)^{n+2}=\left( 2^{n+1}\right)^{n+1}\)
\(则\left( 2^{n+2}\right)^n+\left( 2^{n+2}\right)^n=\left( 2^{n+1}\right)^{n+1}\)
\(则\left( 2^n\right)^{n+2}+\left( 2^n\right)^{n+2}=\left( 2^{n+1}\right)^{n+1}\)
\(若n\left( n+2\right)=ab{,}\ \ \ \left( n+1\right)\left( n+1\right)=cd{,}\)
\(则\left( 2^a\right)^b+\left( 2^b\right)^a=\left( 2^c\right)^d=\left( 2^d\right)^c\)
\(则\left( 2^a\right)^b+\left( 2^a\right)^b=\left( 2^c\right)^d=\left( 2^d\right)^c\)
\(则\left( 2^b\right)^a+\left( 2^b\right)^a=\left( 2^c\right)^d=\left( 2^d\right)^c\)

朱明君

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