Given a = 2x-y, B = x + y, calculate a ^ 2-2ab

Given a = 2x-y, B = x + y, calculate a ^ 2-2ab

A^2-2AB
=A（A-2B）
=(2x-y)(2x-y-2(x+y))
=(2x-y)(-3y)
=3y^2-6xy

Given a = 2x + y, B = 2x-y, calculate A2-B2

A2-B2=（2x+y）2-（2x-y）2=（4x2+4xy+y2）-（4x2-4xy+y2）=4x2+4xy+y2-4x2+4xy-y2=8xy．

A = 2x + y B = 8x-y calculate the square of a - the square of B

A²-B²=(A+B)(A-B)=10x(2y-6x)=20xy-60x²

It is proved that 58-1 solution is divisible by two integers between 20 and 30
The process also needs to be improved~
It's five times eight minus one

8 times of 5 minus 1 = (5 * 5 * 5 * 5 + 1) (5 * 5 + 1) (5 * 5-1)
5*5+1=26
5*5-1=24

Explain that 5 ^ 8-1 can be divided by integers between 20 and 30
The title is as above~
Urgent~
Let's help~

5^8-1
=(5^4+1)(5^4-1)
=(5^4+1)(5^2+1)(5^2-1)
=(5^4+1)*26*24
It's divisible by 24 and 26

If 24a 2 + 1 = B 2, we prove that there is and only one of a and B can be divisible by 5

Since b2-24a2 = 1, it is obvious that a and B cannot be divisible by 5. It is proved that a and B cannot be divisible by 5. If a and B cannot be divisible by 5, then A2 and B2 are of 5K ± 1 type. If one of A2 and B2 is of 5K + 1 type and the other is of 5k-1 type, then A2 + B2 can be divisible by 5. From 24a2 + 1 = B2, 25a2 + 1 = A2 + B2. However, 25a2 + 1 cannot be divisible by 5, so A2 and B2 cannot be one of 5K + 1 type and the other of 5 type If A2 and B2 are both 5K + 1 or 5k-1, then A2-B2 can be divisible by 5, and A2-B2 = - (23a2 + 1). Investigate the single digit of 23a2 + 1: from the above table, we can see that the single digit of 23a2 + 1 does not have 0 or 5, so 23a2 + 1 cannot be divisible by 5, so A2-B2 cannot be divisible by 5. Therefore, A2 and B2 cannot be both 5K + 1 or 5k-1. Then a and B have one and only one can be divisible by 5

Verification: three numbers can be found in any five numbers, and the sum can be divided by three, while three numbers may not be found in four integers, and the sum can be divided by three
I'm in a hurry
Ha ha. I forgot to finish the typing. It's divided by 3

Let the remainder of any five integers divided by 3 be A1, A2, A3, A4, A5 respectively, because the remainder of dividing by 3 may be 0, 1, 2. If there is at least one remainder of 0, 1, 2, then the sum of three integers with the remainder of 0, 1, 2 can be divided by 3. If there is no remainder of 0, 1, 2, then there are three integers according to the drawer principle

Let n be an integer, then all even numbers can be expressed as () and all odd numbers can be expressed as () divisible by 5
Let n be an integer, then all even numbers can be expressed as (), all odd numbers can be expressed as (), the number that can be divided by 5 can be expressed as (), the number that can be divided by 3, and the number of 1 can be expressed as ()

2N 2N-1 5N N/3+1

Let n be a positive integer and prove that n Λ 5-N can be divided by 30

N ^ 5-N = (n-1) n (n + 1) (n ^ 2 + 1), 30 = 2x3x5, because there must be multiples of 2 and 3 in n-1, N, N + 1, we only need to prove that n ^ 5-N can be divisible by 5. When n-1, N, n + 1 has multiples of 5, it obviously holds. When n-1, N, N + 1 has no multiples of 5, then n = 5x + 2 or 5x-2 (x is an integer), then n ^ 2 + 1 = 25X ^ 2 ± 20x + 5 can

The following set (1) a set of all numbers divisible by 3

{XIX = 3N, n is an integer}