Easy. By Yitang Zhang's Theorem, (to be completed soon), there are infinitely many pairs of prime numbers that differ by 2. Therefore, there are infinitely many prime numbers taking both the forms of 4k+1 and 4k+3. End of proof.
【在 b***y 的大作中提到】 : Easy. : By Yitang Zhang's Theorem, (to be completed soon), there are infinitely many : pairs of prime numbers that differ by 2. Therefore, there are infinitely : many prime numbers taking both the forms of 4k+1 and 4k+3. : End of proof.
【在 b***y 的大作中提到】 : Easy. : By Yitang Zhang's Theorem, (to be completed soon), there are infinitely many : pairs of prime numbers that differ by 2. Therefore, there are infinitely : many prime numbers taking both the forms of 4k+1 and 4k+3. : End of proof.
x*p
82 楼
只是应I63的要求一证,没细看每一楼的贴子。
【在 m**x 的大作中提到】 : it's virtually the same as the proof I gave in the update of floor#20. and : you didn't give the proof for infinity number of 4k+1 primes
N*N+1 does not contain prime factor of the form 4k+3. Proof: Let M = N*N + 1 and assume that M contains a prime factor p of the form 4k+3. Then we have the following two observations. (1) gcd(p,N) = 1. According to Fermat's little theorem, N^(p-1) = 1 (mod p) (2) p|M = N*N + 1, then N*N = N^2 = -1 (mod p), then N^4 = 1 (mod p) Now both N^(p-1) and N^2 are equal to 1 (mod p), let k = gcd(4, p-1) and we have N^k = 1 (mod p) Since p has form 4k+3, the p-1 = 4k+2 and thus k = gcd(4, 4k+2) = 2, then we get N^2 = 1 (mod p). But N^2 = -1 (mod p) and thus 1 = -1 (mod p). Then we get p = 2 which is not in the form of 4k+3. We get a contradiction. Done.
O*2
88 楼
有了楼上的结果,我们可以证明4k+1型的素数有无穷多个了。 假设只有有限个4k+1型的素数p_1, ..., p_k,我们考虑 N = 2*p_1*p_2*...*p_k,再 假设 M = N^2 + 1. 我们发现M是个4k+1型的数,且不能被任何一个4k+1型的素数p_i整 除。那么如果M是素数,我们就找到一个新的4k+1型的素数,推出矛盾。如果M不是素数 ,根据87楼的结论,所有M的素因子都是4k+1型的素数,所以M必有一个新的素因子是4k +1型的,依然推出矛盾。
m*x
89 楼
sorry, what is N?
we we we
【在 O********2 的大作中提到】 : N*N+1 does not contain prime factor of the form 4k+3. : Proof: Let M = N*N + 1 and assume that M contains a prime factor p of the : form 4k+3. Then we have the following two observations. : (1) gcd(p,N) = 1. According to Fermat's little theorem, : N^(p-1) = 1 (mod p) : (2) p|M = N*N + 1, then N*N = N^2 = -1 (mod p), then N^4 = 1 (mod p) : Now both N^(p-1) and N^2 are equal to 1 (mod p), let k = gcd(4, p-1) and we : have N^k = 1 (mod p) : Since p has form 4k+3, the p-1 = 4k+2 and thus k = gcd(4, 4k+2) = 2, then we : get N^2 = 1 (mod p). But N^2 = -1 (mod p) and thus 1 = -1 (mod p). Then we
O*2
90 楼
N is any positive integer greater than 1.
【在 m**x 的大作中提到】 : sorry, what is N? : : we : we : we
m*x
91 楼
then why gcd(p,N) = 1?
【在 O********2 的大作中提到】 : N is any positive integer greater than 1.
O*2
92 楼
p is a prime, p divides M = N*N + 1, then p does not divide N. So gcd(p,N)=1 .
【在 m**x 的大作中提到】 : then why gcd(p,N) = 1?
m*x
93 楼
I must have forgotten all the modulus stuff... one last question, how do you get: both N^(p-1) and N^2 are equal to 1 (mod p), let k = gcd(4, p-1) and we have N^k = 1 (mod p)
=1
【在 O********2 的大作中提到】 : p is a prime, p divides M = N*N + 1, then p does not divide N. So gcd(p,N)=1 : .
O*2
94 楼
We know, if k = gcd(a,b), then we can find some integers x and y, such that ax + by = k. Since k = gcd(4, p-1), then we can find x and y, such that 4x + (p-1)y = k Then N^k = N^(4x + (p-1)y) = (N^4)^x * (N^(p-1))^y = 1^x * 1^y = 1 (mod p)
【在 m**x 的大作中提到】 : I must have forgotten all the modulus stuff... : one last question, how do you get: : both N^(p-1) and N^2 are equal to 1 (mod p), let k = gcd(4, p-1) and we : have N^k = 1 (mod p) : : =1
m*x
95 楼
thx. admire
that
【在 O********2 的大作中提到】 : We know, if k = gcd(a,b), then we can find some integers x and y, such that : ax + by = k. : Since k = gcd(4, p-1), then we can find x and y, such that : 4x + (p-1)y = k : Then N^k = N^(4x + (p-1)y) = (N^4)^x * (N^(p-1))^y = 1^x * 1^y = 1 (mod p)
【在 O********2 的大作中提到】 : N*N+1 does not contain prime factor of the form 4k+3. : Proof: Let M = N*N + 1 and assume that M contains a prime factor p of the : form 4k+3. Then we have the following two observations. : (1) gcd(p,N) = 1. According to Fermat's little theorem, : N^(p-1) = 1 (mod p) : (2) p|M = N*N + 1, then N*N = N^2 = -1 (mod p), then N^4 = 1 (mod p) : Now both N^(p-1) and N^2 are equal to 1 (mod p), let k = gcd(4, p-1) and we : have N^k = 1 (mod p) : Since p has form 4k+3, the p-1 = 4k+2 and thus k = gcd(4, 4k+2) = 2, then we : get N^2 = 1 (mod p). But N^2 = -1 (mod p) and thus 1 = -1 (mod p). Then we