Prove that n 2 − 1 is divisible by 8 for any odd integers n.Below...

daktielti

daktielti

Answered

2022-07-15

Prove that n 2 1 is divisible by 8 for any odd integers n.
Below is my proof and I am confused about a few points. I am not sure the final lines are correct as I know that showing 2,4 are factors of 4 k 2 + 4 k is enough to prove that it is divisible by 8 and I have looked at some other examples. In an example of 30, both 3 and 6 are factors of 30 but 30 is not divisible by 18. But I am stuck on how to modify this proof to be complete.
To prove this statement, I intend to use direct proof. Since n has to be an odd integer as prescribed in the statement, by the definition of odd numbers, ( 2 k + 1 ) 2 1 must be divisible by 8 where n = 2 k + 1 for some integer k. Next, we expand ( 2 k + 1 ) 2 1 to be 4 k 2 + 4 k + 1 1, which simplifies to 4 k 2 + 4 k.
First, we use the distributive property to get the following 4 ( k 2 + k ). We let m = k 2 + k and therefore 4 ( k 2 + k ). Hence, we know that 4 k 2 + 4 k must be divisible by 4.
Then, we use the distributive property to factor 4k from the expression to get 4 k ( k + 1 ). By the definition of even and odd number, if k is odd then k + 1 must be even and if k is even then k + 1 is odd. As an integer is Since 2 and 4 are common factors of 4 k 2 + 4 k, which means 8 must also be a factor of 4 k 2 + 4 k.
n 2 1 is therefore divisible by 8 where n = 2 k + 1 for some integer k. Therefore, we have proven that n 2 1 is divisible by 8 for any odd integers n.

Answer & Explanation

Ashley Parks

Ashley Parks

Expert

2022-07-16Added 11 answers

Explanation:
Basically, you only need to know the following fact:
n 2 1 = ( n 1 ) ( n + 1 ) .
And above is the multiplication of two consecutive even numbers.
Augustus Acevedo

Augustus Acevedo

Expert

2022-07-17Added 4 answers

Explanation:
To clarify your proof, note that m = k ( k + 1 ) must be an even number, so k ( k + 1 ) = 2 for some integer ℓ. Now, n 2 1 = 4 k ( k + 1 ) = 8 ,, which proves your result.

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