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.

Question

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.

Ashley Parks · Accepted Answer

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 · Accepted Answer

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