Is the square of any odd number minus 1 a multiple of 8? Please explain why

Is the square of any odd number minus 1 a multiple of 8? Please explain why

Let the odd number be 2x + 1
（2X+1）^2-1
Original formula = 4x ^ 2 + 4x + 1-1
=4X^2+4X
A: not necessarily. When the number is 1, the original formula = 0 is not a multiple of 8

Kneel down for the sum of squares of the four digit positive integers, and find the sum of the four digit positive integers that meet the conditions for 50

There are 24 species and 79 992 species

Try to find all the positive integers a: the last digit is 4, and the sum of squares of all digits is not less than a

a=an*10^n+a(n-1)*10^(n-1)+...+a1*10+a0,a0=4
an^2+a(n-1)^2+...+a1^2+16>=a
1)a=4
2) 10A1 + 4 = 13, A1 > = √ 13 + 5 > 8 or A1

Input 50 positive integers randomly from the keyboard, and count the number in which the sum of squares of each digit is equal to 108

Provides an idea, does not need to copy, has not verified
For example, a number 2268 is 4 + 4 + 36 + 64 = 108
Take each number with a loop, square it and add it
X is the number, y is the cumulative number, and I is the number of cycles
for i=1 to len(x)
y=y+mid(x,i,1)*mid(x,i,1)
next
if y=108 then print x
Pay attention to the conversion between numeric type and character type

Find such three different positive integers, they are mutually prime, and the sum of any two numbers can be divided by the third number

First of all, we assume that the largest number of the three numbers is C, then there must be a + B = C, why not a + B = n * C (n > = 2), because C is the largest number, if a + B = 2C = C + C, then there must be a, B, one of which is larger than C! So we get the first relation, a + B = C, named as a form

It is known that 1-1 / 2 + 1 / 3-1 / 4 + 1 / 5 -... - 1 / 1318 + 1 / 1319 = A / B, a and B are coprime. It is proved that a can be divisible by 1979

S=1-1/2+1/3-1/4+1/5-...-1/1318+1/1319=(1+1/2+1/3+1/4+...+1/1319)-2*(1/2+1/4+...+1/1318)=(1+1/2+1/3+1/4+...+1/1319)-(1+1/2+...+1/659)=1/660+1/661+...+1/13191/660+1/1319=1979/(660*1319),1/661+1/1318=197...

If M and N are positive integers and M = 4N, then the least common multiple of M and N is
It's urgent!

M itself

It is proved that all positive integers in the form of 4K + 1 can be expressed as the sum of squares of two positive integers

When k = 2, 4K + 1 = 9
9=8+1=7+2=3+6=4+5
9 cannot be expressed as the sum of the squares of two positive integers

Proving: 1 + n / 2 by mathematical rule
It's in the back

If a (n) = 1 + 1 / 2 + 1 / 3 +. + 1 / 2 ^ n, then a (n + 1) = 1 + 1 / 2 + 1 / 3 +. + 1 / 2 ^ n + 1 / (2 ^ n + 1) + 1 / (2 ^ n + 2) +... + 1 / 2 ^ (n + 1) first prove the left side: when n = 1, a (n) = 1 + 1 / 2 = 3 / 2; obviously, a (1) > = 1 + 1 / 2 let n = K (k > 1, K is a positive integer)

If the last digit of an integer is 0, the integer can be divided by 5
I thought it was a true proposition, but 0 is also an integer, but it is not divisible by 5

The true proposition stipulates that 0 can be divided by any natural number except 0