设x^a+y^b=z^c

朱明君

\(设x^a+y^b=z^c{,}\ \ 其中x{,}y{,}z{,}a{,}b{,}c均为正整数，\)
一，\(\frac{\ nabc}{\ a}=m{,}\ \ \ \frac{nabc}{\ b}＝k{,}\ \ \ \ c+nabc=dv{,}\)
二，\(\frac{\ nab}{\ a}=m{,}\ \ \ \frac{nab}{\ b}＝k{,}\ \ \ \ c+nab=dv{,}\)
\(则\left( xz^m\right)^a+\left( yz^k\right)^b=\left( z^d\right)^v\)


\(实例\)
\(2^3+1^4=3^2\)
\(一，3\times4=12{,}\)\(\ \ \ \ \ \ \left( 2\times3^4\right)^3+\left( 1\times3^3\right)^4=\left( 3^2\right)^7\)
\(二，3\times4\times2=24{,}\)\(\ \ \ \ \ \ \left( 2\times3^8\right)^3+\left( 1\times3^6\right)^4=\left( 3^2\right)^{13}\)
\(三，2\left( 3\times4\times2\right)=48{,}\)\(\ \ \ \ \ \ \left( 2\times3^{16}\right)^3+\left( 1\times3^{12}\right)^4=\left( 3^2\right)^{25}=\left( 3^{10}\right)^5\)
\(\cdots\cdots。\)
勾股数能解\(x^2+y^2=z^5\)，\(x^4+y^4=z^5\)，\(x^2+y^4=z^5\)，\(x^4+y^2=z^5\)
\(x^4+y^4=z^2\)，\(x^2+y^4=z^2\)，\(x^4+y^2=z^2\)

\(3^2+4^2=5^2\)
      \(\left( 3\times5^4\right)^2+\left( 4\times5^4\right)^2=\left( 5^2\right)^5\)
\(5^2+12^2=13^2\)
      \(\left( 3\times13^4\right)^2+\left( 4\times13^4\right)^2=\left( 13^2\right)^5\)
\(\cdots\cdots。\)

朱明君

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