The solution of the equation a (x + m) 2 + B = 0 of X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), then the solution of the equation a (x + m + 2) 2 + B = 0 is______ ．

The solution of the equation a (x + m) 2 + B = 0 of X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), then the solution of the equation a (x + m + 2) 2 + B = 0 is______ ．

∵ the solution of the equation a (x + m) 2 + B = 0 about X is X1 = - 2, X2 = 1, (a, m, B are all constants, a ≠ 0), ∵ the equation a (x + m + 2) 2 + B = 0 is transformed into a [(x + 2) + M] 2 + B = 0, that is, in this equation x + 2 = - 2 or x + 2 = 1, the solution is x = - 4 or x = - 1. So the answer is: X3 = - 4, X4 = - 1

Let prime number P > 3, for all a and B belong to integer, prove: 6p integral division (a * B ^ P-B * a ^ P)

The other formula is f
Obviously 2 | f (parity) 3 | f (enumeration a, B divided by 3)
Now we need to prove: p|f
Fermat's theorem: x ^ P = x mod (P) P is prime
f = a*b - b*a = 0 mod(p)
So p | f

If P is a prime and a is any integer, then a can be divisible by P or P and a are coprime (the greatest common factor of P and a is 1). Why can a be divisible by P, for example, a = 3, P = 2

If P is a prime and a is any integer, then a can be divisible by P or P and a are coprime (the greatest common factor of P and a is 1)
What's your example
It should not be divisible, or it should be coprime
A is divisible by P
or
P and a are coprime

Given that a and B are opposite to each other, C and D are reciprocal to each other, and X is the smallest prime number, try to find the value of X - (a + B + CD) + |a + B-4 | + |3-cd |

a. B is opposite to each other, C and D are reciprocal to each other, and X is the smallest prime number
a+b=0 cd=1 x=2
x—(a+b+cd)+|a+b—4|+|3—cd|
=2-(0+1)+|0—4|+|3—1|
=2+1+4+2
=9

It is proved that if P is an odd prime, then the prime Q which can divide 2 ^ P-1 must be a multiple of 2p plus 1

Let 2 ^ P-1 = a * q, where q is any odd prime number of 2 ^ P-1. Then there is Q ≡ 1 (MODP), thus a * q ≡ 1 (MODP), 2 ^ P-1 ≡ 1 (MODP). Also let q = NP + 1, suppose n ≠ 2m (where n and m are natural numbers), then Q-1 cannot be divisible by 2, then q is even. It is impossible to know from 2 ^ P-1 = a * q! Because 1 cannot be divisible by 2! This is impossible. We get n = 2m, q = NP + 1, that is q = 2MP + 1, That's what we got

Let the integer k, K ≥ 14, p be the largest prime number less than k, P ≥ 3kg4, n be a composite number. It is proved that if n is greater than 2p, then n can divide (n-k)!

Coordinate axis diagram: --- 0.25k--0.5k--0.75k--p -- K -------- 2p -------- n-----------
Suppose q is any prime factor of N, QP, so Q > = k, n-k > = n-q = Q
If n = 3 * q, then n = 3Q > 2p > = 1.5k, Q > = 0.5K, let q = 0.5K + X, then n-k = 0.5K + 3x > Q
If n = h * q, H > = 4, n = HQ > 2p = 1.5k, let q = 1.5k/h + X, where x > 0, then n-k = 0.5K + HX > Q
We can see all the prime factors Q of n

Let p be a prime number greater than 3, and prove that 3 can divide the square of P by 1

The square of P minus 1 = (P + 1) * (p-1)
And P is a prime number, so p must not be divisible by 3: its value must be + 1 or - 1 (i.e. + 2) of the integer multiple of 3. That is to say, its value of + 1 or - 1 must be divisible by 3

It is proved that when p is an odd prime, there are 1 ^ P + 2 ^ P + 3 ^ P + ·· + (p-1) ^ (p-1) congruences with 0 module P

Your title is wrong! It's (p-1) ^ P, otherwise it's not regular!
Using Fermat's law
Because P is prime, then p and 1, 2, 3 And (p-1) are mutually prime,
So there is a ^ (p-1) ≡ 1 (mod p)
So a ^ P ≡ a (mod p)
therefore
The original formula ≡ 1 + 2 + 3 + +（p-1） （mod p）
≡ P (p-1) / 2 (mod p) (∵ P is odd prime, so P-1 is even and divisible by 2)
≡0 （mod p）
If you haven't learned Fermat's law, first understand the rest class, then Baidu "Fermat's law"

It is proved that if a > 1, there is a prime number P such that A

When a = 2. A! + 1 = 3, there is a prime number 32

Proof: if P is prime and P ≡ 1 (MOD 4), then {[(p-1) / 2]!} ^ 2 + 1 ≡ 0 (mod p), please help me,

This is a part of the famous Euler criterion. For any integer 1 & lt; = I & lt; = P-1, there is always a unique integer J, where I * J is divided by P, and the remainder is B. since B is a quadratic non residue of P, I is not equal to j, so 1,2 P-1 is divided into (p-1) / 2 pairs, the product congruence B of each pair, so there is (p-1)! Congruence B ^ ((p-1) / 2)