关于形式为1+2^（2^n）+2^（2^n +1）+.+的和的一个定理2^（2^n+m）和Fermat与Mersenne数联系的一个结果
On a Theorem on sums of the form 1+2^(2^n)+2^(2^n+1)+...+2^(2^n+m) and a result linking Fermat with Mersenne numbers
摘要：
在他的书“250个问题的基本数论“中，W.谢尔宾斯基表明，数字1+2^（2^n）+2^（2^n +1）是可被21整除的;对于n= 1，2，.本文证明了一个类似但更一般的结果，考虑I（n.m）= 1+2^（2^n）+2^（2^n +1）+.+的自然数2^（2 ^n +m）.在定理1中，我们证明了对每一个大于1的奇数N，存在无穷多个自然数n和m使得整数I（n.m）能被N整除.对于给定的N，我们给出了数n和m的显式构造。例如，当N=31，并且n=4k和m=94+ 124 i时，数字I（n，m）可被31整除。对于N=（31）（7）=217提供了类似的例子。在定理2中，我们证明了一个与梅森数有关的结果，并给出了三个推论，其中一个是关于费马数的。
英文摘要：
In his book "250 Problems in Elementary Number Theory", W.Sierpinski shows that the numbers 1+2^(2^n)+2^(2^n+1) are divisible by 21; for n=1,2,.... In this paper, we prove a similar but more general result.Consider the natural numbers of the form I(n.m)= 1+2^(2^n)+2^(2^n+1)+...+2^(2^n+m).In Theorem 1 we prove that for every odd integer N greater than 1, there exist infinitely many natural numbers n and m such that the integers I(n.m) are divisible by N. We give an explicit construction of the numbers n and m, for a given N. As an example, when N=31, and with n=4k and m=94+124i, the numbers I(n,m) are divisible by 31. A similar example is offered for N=(31)(7)=217. In Theorem 2, we prove a result pertaining to Mersenne numbers.There are also three Corollaries in this work, one of which deals with Fermat numbers.
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