$b^2 + 1$型素数的无穷多个
An Infinitude of Primes of the Form $b^2 + 1$
摘要：
如果b^2 + 1是素数，那么b一定是偶数，因此我们考察4u^2 + 1的形式。我们不研究这种形式的素数，而是研究合数，其中本文的主要定理建立了，如果4u^2 + 1是合数，则u属于一个集合，其元素是使得u^2 + t^2 = n（n + 1）的那些u，其中t有一个由u的值确定的上界。这将4u^2 + 1形式的合数与可表示为两个平方和等于两个连续整数乘积的数联系起来。高斯关于两个平方和表示数的平均数得到的一个结果可以用来证明u的无穷序列使得u^2 + t^2 = n（n + 1）是不可能的。这就意味着不可能有无穷序列的复合体，因此也就不可能有无限个b^2 + 1形式的素数。
英文摘要：
If b^2 + 1 is prime then b must be even, hence we examine the form 4u^2 + 1. Rather than study primes of this form we study composites where the main theorem of this paper establishes that if 4u^2 + 1 is composite, then u belongs to a set whose elements are those u such that u^2 + t^2 = n(n + 1), where t has a upper bound determined by the value of u. This connects the composites of the form 4u^2 + 1 with numbers expressible as the sum of two squares equal to the product of two consecutive integers. A result obtained by Gauss concerning the average number of representations of a number as the sum of two squares is then used to show that an infinite sequence of u for which u^2 + t^2 = n(n + 1) is impossible. This entails the impossibility of an infinite sequence of composites, and hence an infinitude of primes of the form b^2 + 1.
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