命题　设 a,b,d∈ N+ ,且 ( d,b) =1 ,则首项为 a,公差为 d的等差数列中一定有无穷多项是 b的自然数次幂 .证明　∵　 an =a + ( n - 1 ) d,∴　 an =bk　 　 d| ( bk - a) .构造数列 :b- a,b2 - a,… ,bm- a,… ,则由抽屉原理 ,必有 m1m2 ,使得( bm2 - a) - ( bm1- a)≡ 0　 If the proposition is a, b, d ∈ N +, and (d, b) = 1, there must be infinitely many items in the arithmetic sequence with the first item a and the tolerance d as the natural power of b. Proof ∵ an = a + (n - 1) d, ∴ an = bk d | (bk - a). Constructing series: b- a, b2 - a, ..., bm- a, ..., then by drawer principle, there must be m1m2, (bm2 - a) - (bm1 - a) ≡ 0