Prove that the difference between the squares of any two odd numbers is a multiple of eight?

Question

You may also assume that any odd number can be written in the form 2r-1, where r is an integer

Answer 1

Let the two numbers be:
2m + 1 and 2n + 1 where m,n are integers
Then D = (2m+1)^2 - (2n+1)^2
= 4m^2 + 4m + 1 - (4n^2 + 4n + 1)
= 4m^2 - 4n^2 + 4m - 4n
= 4(m - n)(m + n ) + 4(m - n)
= 4(m - n)(m + n + 1)

which is divisible by 4. All that is left is that we show that one of (m-n) and (m+n+1) is divisible by 2.

If m and n are both even then m-n is even and we are done.
If m and n are both odd then m-n is even and we are done.
If one of m,n is odd and the other is even (call m odd and n even without loss of generality) then:
let m = 2k+1 and n = 2j for some integers j, k.
Then (m - n + 1) = 2k+1 + 2j -1 = 2k + 2j = 2(k+j) and we are done.

Answer 2

Let the odd numbers be 2r+1 and 2s+1, where r and s are positive whole numbers

(2r-1)^2 - (2s-1)^2
= 4r^2 - 4r + 1 - 4s^2 + 4s - 1
= 4(r^2 - s^2) - 4(r-s)
= 4(r-s)(r+s) - 4(r-s)
= 4(r-s)(r+s-1)

We know that the difference of the square of 2 odd numbers will be a multiple of 4 from above.

If r and s are both odd or both even, (r-s) becomes even, i.e. a multiple of 2.

If r and s are such that one is even and the other odd, then (r+s-1) is always even, i.e. a multiple of 2.

Hence, (2r-1)^2 - (2s-1)^2 = 4(r-s)(r+s-1) is always a multiple of 8.

Answer 3

Lets let the first number be 2m-1 and the second be 2n-1. Then their squares are 4m^2-4m+1 and 4n^2-4n+1

Subtracting them gives 4[(m^2-m)-(n^2-n)]

Now consider m^2-m. we can factor this as m(m-1). Now either m is even or (m-1) is even. So the product m(m-1) MUST be even, since it has an even number as a factor. So m^2-m is even. Similarly n^2-n must be even as well. Since the difference of two even numbers is even, we know that everything inside the brackets above must be even. That is:

[(m^2-m)-(n^2-n)] must be even
[(m^2-m)-(n^2-n)] = 2p for some integer p

So the dfference of two squares may be written as 4(2p) or 8p
So the difference MUST be divisble by 8

Answer 4

Let x and y be odd numbers. Then

x = (2k - 1)
y = (2z - 1)
for k,z integers.

(I want to prove that x^2 - y^2 = 8a, for some integer a).

x^2 - y^2 = (2k - 1)^2 - (2z - 1)^2
(4k^2 - 4k + 1) - (4z^2 - 4z + 1)

4k^2 - 4k + 1 - 4z^2 + 4z - 1
4k^2 - 4k - 4z^2 + 4z
4(k^2 - k - z^2 + z)
4( k(k-1) - z(z-1) )

At this point, note that ANY number multiplied by its previous number is even, because one of them has to be even. The difference between two even numbers is an even number, therefore, k(k-1) - z(z-1) itself has to be even.

So k(k-1) - z(z-1) is a multiple of 2, and can be expressed as 2b. i.e.

4( k(k-1) - z(z-1) ) = 4(2b) = 8b

Showing that it's a multiple of 8.

Answer 5

Okay let r and s be two integers.
Then, (2r-1)^2-(2s-1)^2 = 4r^2-4r+1-4s^2+4s-1 = 4r^2-4r-4s^2+4s= 4(r^2-r-s^2+s) = 4(r(r-1)) + 4(s(1-s)).

Now, r is either even or odd.
Now if r is even , it satifies the condtion, since every even number is a multiple of 2.
Now, if r is odd, that would make r-1 even, so it stil satifies the condition.

And same arguement can be used with s.

Answer 6

ok so for any entire volume n, the expression: 2n+a million is consistently an abnormal volume. the subsequent abnormal volume might want to then be 2n + 3 because each abnormal volume is two remote from the subsequent one. hence: (2n + 3)^2 - (2n + a million)^2 = (4n^2 + 12n + 9) - (4n^2 + 4n + a million) = 4n^2 + 12n + 9 - 4n^2 - 4n - a million= 8n - 8 = 8(n -a million) which signifies that no be counted what, the answer will be flippantly divisible with the aid of 8 because we declared than change into an entire volume so: 8(n-a million)/8 = n -a million it continues to be an entire volume.

Answer 7

(2r-1)^2 = 4r^2 - 4r + 1

(2s-1)^2 = 4s^2 - 4s + 1

Subtract and get 4r^2 - 4r - 4s^2 + 4s

= 4(r^2 - r - s^2 + s)

= 4(r(r-1) - s(s-1))

But r and r-1 must be odd and even, or even and odd, so their product is even and a multiple of 2 (same for s and s-1)

Since it already is a multiple of 4, it's now a multiple of 8.

Answer 8

(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)
8n. for any n is an integer, 8n is divisible by 8. for all other odd integers, we can get there by extending this. since it valid for 2 adjacent odd jntegers such as 2n-1 & 2n+1, it will be valis for exam ple for 2n-1 & 2n-3 & therefore for 2n-3 & 2n+1 sinc ethe sum of numbers divisible by 8 will also be divisible by 8. This can be exrended to all odd integers since you will have sums of numbers divisible by 8.

Answer 9

There is nothing in your question that says the numbers have to be different, so if both odd numbers are the same then your statement is false.

Answer 10

how can that be? the difference between 3 squared and 5 squared is 9