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Why is the greatest common divisor multiplied by the least common multiple the product of these two numbers?
Two, except for the greatest common divisor 13, you only need to find the product of the greatest common divisor 13 and the greatest common divisor 13
It is known that the sum of squares of two natural numbers is 900, and the product of their greatest common divisor and least common multiple is 432
Let MT, NT, t be the greatest common divisor of the two numbers, m, n be coprime, let m > n. (MT) &# 178; + (NT) &# 178; = 900t × (MNT) = 432, then (M & # 178; + n & # 178;) T & # 178; = 900 (1) MNT & # 178; = 432 (2) (1) / (2) (M & # 178; + n & # 178;) / (MN) = 25 / 12, then 12m & # 17
It is proved that there are infinitely many prime numbers of type 4k-1
Proof: the counter proof assumes that there are finite primes of type 4k-1, no matter n, P1, P2 PN let a = (P1 * P2 *...) PN) ^ 2 + 2 due to (P1 * P2 * If a is a prime, then a is a prime of type 4k-1, and there is no contradiction among the N primes. If a is a composite number, it is obvious that there must be at least one prime of type 4k-1 in the prime factor of A. otherwise, if a is a prime of type 4k-1, then it is easy to see that B is not in the N primes, The hypothesis is not true. That is to say, there are infinite primes of type 4k-1. What the answer of Zheng Bi upstairs proves is 4K + 1 instead of 4k-1
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