关于方程sum from k=0(x-2^(1)5~αq^k)~r=sum from k=1 to n(x+2^(1)5~αq^k)~r

ON THE DIOPHANTINE EQUATION sum from k=0(x-2^(1)5~αq^k)~r=sum from k=1 to n(x+2^(1)5~αq^k)~r

本文证明了,q=p_1p_2…p_τ=10β+3型奇数,p_1,p_2,…,p_τ是不同素数,n,x,s,α,r为正整数时,方程sum from k=o to n(x-2~s5~αq^K)~r=sum from k=1 to n(x+2~S5~αq^K)~r,仅有正整数解r=1,x=2~s5~αqn(n+1)和r=2,x=2^(s+1)5~αqn(n+1).

In this paper, we proved following results: let q= p1p2 pi= 10B+3 be odd, p1, p2, , pi, be distinct prime, n, x, e , a, r, be positive integer, the equation has no positive integer solutions,