不定方程x^a+y^b=z^c的研究

朱明君

(第1题)
\(不定方程x^n+y^{\left( n+1\right)}=z^{\left( n+2\right)}的通解公式\)

\(一，设n为奇数，\)
\(\frac{n\left( n+2\right)+1}{2}=m，\)
\(则\left( 2^m\right)^n+\left( 2^{m-\left( \frac{n+1}{2}\right)}\right)^{n+1}=\left( 2^{m-n}\right)^{n+2}\)
\(实例n=17,\)
X^17+y^18=z^19
\(\frac{17\times19+1}{2}=162，\frac{18}{2}=9，\)
\(\left( 2^{162}\right)^{17}+\left( 2^{\left( 162-9\right)}\right)^{18}=\left( 2^{\left( 162-17\right)}\right)^{19}\)
\(实例n=2023,\)
x^2023+y^2024=z^2025
\(\frac{2023\times2025+1}{2}=2048288{,}\ \ \ \ \frac{2024}{2}=1012\)
\(\left( 2^{2048288}\right)^{2023}+\left( 2^{\left( 2048288-1012\right)}\right)^{2024}=\left( 2^{\left( 2048288-2023\right)}\right)^{2025}\)

\(二，设n为偶数，\)
\(则\left( \left( 2^{n\left( n+2\right)}-1\right)^{n+2}\right)^n+\left( \left( 2^{n\left( n+2\right)}-1\right)^{n+1}\right)^{n+1}=\left( \left( 2\times\left( 2^{n\left( n+2\right)}-1\right)\right)^n\right)^{n+2}\)
\(实例n=2,\)
\(\left( \left( 2^8-1\right)^4\right)^2+\left( \left( 2^8-1\right)^3\right)^3=\left( 2\times\left( 2^8-1\right)^2\right)^4\)
\(实例n=10,\)
\(\left( \left( 2^{120}-1\right)^{12}\right)^{10}+\left( \left( 2^{120}-1\right)^{11}\right)^{11}=\left( 2\times\left( 2^{120}-1\right)^{10}\right)^{12}\)

\(三，设n为大于等于2的正整数，\)
\(则\left( \left( 2^n-1\right)^n\right)^{n-2}+\left( \left( 2^n-1\right)^{n-1}\right)^{n-1}=\left( 2\times\left( 2^n一1\right)^{n-2}\right)^n\)
\(实例n=2,\)


(第2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3题)
\(设\ x^a+y^b=z^c{,}\ 其中\ x{,}y{,}z{,}a{,}b{,}c{,}n{,}为正整数，\)
\(若a是nb的倍数，则\left( xz\right)^a+\left( yz^n\right)^b=z^{a+c}\)