It is proved by factorization that the square of any odd number minus 1 is a multiple of 8

Let this odd number be 2N-1 (n > = 2), then the square minus 1 of this number is expressed as: (2n-1) ^ 2-1 = [(2n-1) + 1] * [(2n-1) - 1] = 2n (2n-2) = 4 * n * (n-1) if n is even, then there must be an X such that n = 2 * x, the original formula = 8 * x (n-1); if n is odd, then (n-1) is even, there must be a y such that (n-1) = 2 * y, the original formula = 8

The square of an odd number subtracts the square of another odd number which is two times smaller than it. The results are all multiples of 8. How to express them in the equation (let n be an odd number)
The following formula is given
3^-1^=8=8×1
5^-3^=16=8×2
7^-5^=24=8×3
9^-7^=32=8×4
Observe the series of equations, analyze their laws, and express the laws with equations. (^ This is the square sign)

All of the above are correct

It is proved that if M and N are positive even or odd numbers, then the square of M minus the square of N must be a multiple of 4
Thank you very much!

m=2x
n=2y
M ^ 2-N ^ 2 = 4 (x ^ 2-y ^ 2) is a multiple of 4
m=2x+1
n=2y+1
M ^ 2-N ^ 2 = 4 (x ^ 2 + X-Y ^ 2-y) is a multiple of 4

p: Integer a can be divided by 5, Q: what is the condition of Q when the last digit of integer a is 5

When the last digit of a is 5, a can be divided by 5, so Q can deduce P;
If a can be divided by 5, then the last digit of a is 0 or 5, so p cannot deduce Q,
It can be seen that P is a necessary and insufficient condition for Q

p: Integer a can be divided by 5, Q: the last digit of integer a is 5. Some students expressed the proposition PVQ as an integer that can be divided by 5. The number of digits must be 5 or 0?
Is this correct? Why? If not, how

Mathematics: the number of digits that can be divided by 5 is 0 or 5. The number of digits that can be divided by 5 must be 5 or 0. Obviously, the last digit of integer a is 5 or the last digit of integer a is 0, that is, the last digit of QV integer a is 0. What do you mean by P. "integer a can be divided by 5"

Can you explain the reason that the last digit is an integer of 0 or 5, which can be divided by 5?

The number whose last digit is 0 can be written as 10N. N is an integer,
If the last digit is 5, it can be written as (10N + 5), and N is an integer,
obviously,
10N △ 5 = 2n, 2n is an integer, 10N can be divided by 5
(10N + 5) △ 5 = 2n + 1, 2n + 1 is an integer, and (10N + 5) can be divided by 5
Therefore, an integer whose last digit is 0 or 5 must be divisible by 5

If you give five integers, you must choose three of them so that their sum can be divided by three

3.6.9.12.15
The sum of multiples of three plus multiples of three should also be multiples of three

From any five integers, we can find three, so that the sum of these three numbers can be divided by 3. Why?

We consider that there are only three possibilities for the remainder of these five numbers divided by 3: (0 means divisible) 0,1,2. We classify five numbers according to the remainder of their division by 3. If three or more of these five numbers are equal to the remainder of their division by 3, then it is obvious that the sum of three of them can be divisible by 3

How to prove that the sum of the digits of an integer can be divided by 3, and it can be divided by 3
I'm a little confused, for example
126's tens add up to 1 + 2 + 6 = 9, which can be divisible by 3, so it can be divisible by 3
Add 30 points

It's a bit complicated to express. Take the four digit number as an example, let the four digit number be a, B, C and D respectively
Then, this number is 1000A + 100b + 10C + D
The sum of each digit is a + B + C + D
If a + B + C + D is divisible by 3
Then this number can be expressed as 1000A + 100b + 10C + D
=（a+b+c+d）+999a+99b+9c
=（a+b+c+d）+3（333a+33b+3c）
So, it can also be divided by 3
The other n-digit numbers are the same as the others

Prove: 58 - 1 can be divided by two integers between 20 and 30

∵ 58-1 = (54 + 1) (54-1), = (54 + 1) (52 + 1) (52-1), = (54 + 1) × 26 × 24. 58-1 can be divisible by 26 and 24 integers between 20 and 30

It is proved by factorization that the square of any odd number minus 1 is a multiple of 8