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).
文摘本文证明了,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).
