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If a (M + 1, n-1,3), B (2m, N, m-2n), C (M + 3, n-3,9), M + n =? In addition, I want to know how to calculate the length of line segment in the solid coordinate system?
A (z.x.c) B (a, s, d) AB = radical (A-Z) ^ + (s-x) ^ + (d-c) ^ a (M + 1, n-1,3). B (2m, N, m-2n). C (M + 3, n-3,9), collinear 2m / M + 1 = n / n-1 = m-2n / 32m / M + 3 = n / n-3 = m-2n / 9m-2n / 3 = 3 (m-2n / 9) n / n-1 = 3 (n / n-3) 2m / M + 1 = 3 (2m / M + 3) M = 0 n
The prime::: m is a positive integer
1）
hypothesis
M is a composite number
Let m = P * t (P and T are integers, P > 1)
Yi Zhi p
Given that M > 1, M is an integer and M is divided by (m-1)! + 1, we prove that M must be a prime number
It is proved that if M is not a prime number and the minimum prime factor of M is p (P > 2), obviously, M > = P ^ 2
Then M-1 > = P ^ 2-1 = (p-1) (P + 1) > = P + 1 > p
Obviously P | (m-1)!
According to the meaning m | (m-1)! + 1, there is obviously P | (m-1)! + 1
=>P | ((m-1)! + 1 - (m-1)!) = > P | 1 contradiction
So if the converse is not true, the original proposition is true
That's it!
This is actually Fermat's theorem
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