不定方程求解公式

朱明君

(第1题)
\(设正整数n大于等于1，\)
\(则2^n+2^n=2^{n+1}\)
\(则\left( 2^{n+2}\right)^n+\left( 2^{n+2}\right)^n=\left( 2^{n+1}\right)^{n+1}\)
\(则\left( 2^n\right)^{n+2}+\left( 2^n\right)^{n+2}=\left( 2^{n+1}\right)^{n+1}\)
\(则\left( 2^{n+2}\right)^n+\left( 2^n\right)^{n+2}=\left( 2^{n+1}\right)^{n+1}\)
\(若n=ab{,}\)
\(则2^n=\left( 2^a\right)^b=\left( 2^b\right)^a\)
\(若n+1=cd{,}\)
\(则2^{n+1}=\left( 2^c\right)^d=\left( 2^d\right)^c\)
\(则\left( 2^{n+1}\right)^{n+1}=\left( \left( 2^{n+1}\right)^c\right)^d=\left( \left( 2^{n+1}\right)^d\right)^c\)
(第2题)
\(设正整数a\ge2{,}\ \ n\ge1{,}\ \ \ \ \)
\(则\left( a^n-1\right)^n+\left( a^n-1\right)^{n+1}=\left( a\times\left( a^n-1\right)\right)^n\)
\(则\left( \left( a^{n\left( n+2\right)}-1\right)^n\right)^{n+2}+\left( \left( a^{n\left( n+2\right)}-1\right)^{n+1}\right)^{n+1}=\left( \left( a\times\left( a^{n+2}-1\right)\right)^n\right)^{n+2}\)
\(则\left( \left( a^{n\left( n+2\right)}-1\right)^{n+2}\right)^n+\left( \left( a^{n\left( n+2\right)}-1\right)^{n+1}\right)^{n+1}=\left( \left( a\times\left( a^{n+2}-1\right)\right)^{n+2}\right)^n\)
\(则\left( \left( a^{n\left( n+2\right)}-1\right)^{n+2}\right)^n+\left( \left( a^{n\left( n+2\right)}-1\right)^{n+1}\right)^{n+1}=\left( \left( a\times\left( a^{n+2}-1\right)\right)^n\right)^{n+2}\)
(第3题)
\(设正整数m大于等于1，n大于等于1，\)
\(则\left( 2^m\right)^n+\left( 2^m\right)^n=2^{mn+1}\)
\(则\left( 2^n\right)^m+\left( 2^n\right)^m=2^{mn+1}\)
\(则\left( 2^n\right)^m+\left( 2^m\right)^n=2^{mn+1}\)
\(若mn+1=ab{,}\ \ \ \ 则2^{mn+1}=\left( 2^a\right)^b=\left( 2^b\right)^a\)
(第4题)
\(设\ x^a+y^b=z^c{,}\ 其中\ x{,}y{,}z{,}a{,}b{,}c{,}n为正整数，\)
\(则\left( xz^{nb}\right)^a+\left( yz^{na}\right)^b=z^{nab+c}\)

(第5题)
\(设\ x^a+y^b=z^c{,}\ 其中\ x{,}y{,}z{,}a{,}b{,}c{,}n{,}为正整数，\)
\(若a是nb的倍数，则\left( xz\right)^a+\left( yz^n\right)^b=z^{a+c}\)

朱明君

\(设x^2-1=mn，其中正整数x-1=m，x+1=n，a大于等于2，\)
\(则\left( \left( a^{mn}-1\right)^n\right)^m+\left( \left( a^{mn}-1\right)^x\right)^x=\left( \left( a\times\left( a^{mn}-1\right)\right)^m\right)^n\)
\(则\left( \left( a^{mn}-1\right)^n\right)^m+\left( \left( a^{mn}-1\right)^x\right)^x=\left( \left( a\times\left( a^{mn}-1\right)\right)^n\right)^m\)
\(则\left( \left( a^{mn}-1\right)^m\right)^n+\left( \left( a^{mn}-1\right)^x\right)^x=\left( \left( a\times\left( a^{mn}-1\right)\right)^m\right)^n\)

白新岭

朱明君先生玩这个到有一套自己的方法。

朱明君

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