丢番图方程a^x+b^y=c^z的正整数解

Positive Integer of Diophantine Equation a^x+b^y=c^z

设m为正整数,且a=m7-21m5+35m3-7m,b=7m6-35m4+21m2-1,c=m2+1.本文同时利用2个代数数的线性型下界估计以及2个有理数方幂之差的p-adic值的下界估计的一些深入结果,证明了对正整数m≥2.4×109,丢番图方程ax+by=cz仅有正整数解(x,y,z)=(2,2,7).

In our report, let m ∈n,a =m^7 -21m^5 ＋35m^3 -7m,b =7m^6 -35m^4 ＋21m^2 - 1 ,c =m^2 ＋ 1, A deep result of the lower bound for linear forms in two logarithms and the lower bound for the p-adic distance between two powers of rational numbers were used to prove that if m≥2. 4 x 109, the Diophantine equation a^x ＋ b^y = c^z has only one positive integer solution （x,y,z） = （2,2,7）.