2018年10月10日 · 如果只有有限个4k＋1型素数p1...pn，考虑K＝（2p1...pn）^2＋1，K的素因子都是4k＋1型的，又不在p1…pn中，矛盾。 发布于 2018-10-29 09:32 赞同 59 14 条评论

I have proved that $-1$ is not a quadratic residue modulo $4k-1$ and is a quadratic residue modulo $4k+1$. Thus I need to prove that there are infinitely many primes of the form $b^2+1,\ b\in\mathbb{N}$.

2021年9月13日 · （1）除2之外的所有素数均为奇数，都可以表示为4k+1或者4k-1 。 （2）且任何一个数都可以表示为某些素数的乘积形式（哥德巴赫猜想）。 反证法：假设一个4k-1型的数N可以表示为：N=（4k1+1）（4k2+1）....（4kn+1)

Let $p$ be a prime number of the form $4k+1$. Fermat's theorem asserts that $p$ is a sum of two squares, $p=x^2+y^2$. There are different proofs of this statement (descent, Gaussian integers,...). And recently I've learned there is the following explicit formula (due to Gauss): $x=\frac12\binom{2k}k\pmod p$, $y=(2k)!x\pmod p$ ($|x|,|y|<p/2$).

2018年11月27日 · We saw in Chapter 5 that in order for a prime to divide the side length of a primitive right triangle, it has to be of the form \ (4k+1\). It would be extremely surprising, and rather unfortunate, if there were only finitely many such primes. In this chapter we will prove the following theorem: Theorem 6.1.

2017年9月21日 · We can consider this sequence $3, 5, 7, 11, 19, 23, 31, \ldots$, which are all a form of $4k - 1$. So, we start by considering the positive integers of the form $n=4k-1$ and their possible prime divisors. Show that every prime divisor of $4k - 1$ is odd. Show that an odd number is either of the form $4k - 1$ or $4k + 1$.

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2019年3月11日 · 如何证明4k+1型素数只能表示成一对正整数的平方和？ 如果是ac+bd，那么a^2+b^2+c^2+d^2-2ac-2bd是一个小于2p的p倍数，要命的是这个值还是个偶数，并且这个值也可以写成 (a-c)^2+ (b-d)^2，那么这个数就是0。 就有a=c,b=d，矛盾. 推广 . 参考资料 .

2016年12月28日 · 无水不欢：怎么证明4k+1素数有无穷个？ 用初等证明？ 看来是竞赛生， 狄利克雷定理 了解一下。 下面是柯召孙琦《数论讲义》上册. Dirichlet 定理，但是我想题主想要一个 …

2020年1月31日 · All Prime numbers except 2 are divided into numbers that are represented as 4k+1 and numbers that are represented as 4k-1, where k is some integer. Fermat's theorem on primes states that primes of the first group are always represantable as the sum of two squares, whereas primes of the second group are never representable as the sum of two squares.