一个素数和一个素数的k次方和问题

ON THE SUM OF A PRIME AND k-POWER OF A PRIME

设k≥ 2 ,Hk 表示一个正整数n的集合 ,使对任意的正整数q ,同余方程a +bk≡n(modq)在模q的既约剩余系中有解a ,b .Ek(x)表示n≤x ,n∈Hk,但不能表成p1+p2 k=n的数的个数 ,则在GRH下有Ek(x) x1-2h(k)4 k- 1 +ε,这里h( 2 ) =316 ;k>2 ,h(k) =4k-12× ( 3× 4k -2 +1)k.

Let k≥2, H-k deuote the set of all numbers n such that a+b+k≡n(mod q) has solutions in reduced residues a, b(mod q) for any tuteger q. Let E-k(x) be the number of all n≤x, n∈H-k which cannot be written as p-1+pk 2=n. Then assuming GRH, E-k(x)x 1-2h(k)4k-1+ε, where h(2)=316, k≥2, h(k)=4 k-12×(3×4 k-1+1)k.