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 $4{k}^{2}+4k$ 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, $(2k+1{)}^{2}-1$ must be divisible by 8 where $n=2k+1$ for some integer k. Next, we expand $(2k+1{)}^{2}-1$ to be $4{k}^{2}+4k+1-1$, which simplifies to $4{k}^{2}+4k$.

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 $4{k}^{2}+4k$ must be divisible by 4.

Then, we use the distributive property to factor 4k from the expression to get $4k(k+1)$. 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 $4{k}^{2}+4k$, which means 8 must also be a factor of $4{k}^{2}+4k$.

${n}^{2}-1$ is therefore divisible by 8 where $n=2k+1$ for some integer k. Therefore, we have proven 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 $4{k}^{2}+4k$ 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, $(2k+1{)}^{2}-1$ must be divisible by 8 where $n=2k+1$ for some integer k. Next, we expand $(2k+1{)}^{2}-1$ to be $4{k}^{2}+4k+1-1$, which simplifies to $4{k}^{2}+4k$.

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 $4{k}^{2}+4k$ must be divisible by 4.

Then, we use the distributive property to factor 4k from the expression to get $4k(k+1)$. 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 $4{k}^{2}+4k$, which means 8 must also be a factor of $4{k}^{2}+4k$.

${n}^{2}-1$ is therefore divisible by 8 where $n=2k+1$ 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

Expert

2022-07-16Added 11 answers

Explanation:

Basically, you only need to know the following fact:

${n}^{2}-1=(n-1)\cdot (n+1).$

And above is the multiplication of two consecutive even numbers.

Basically, you only need to know the following fact:

${n}^{2}-1=(n-1)\cdot (n+1).$

And above is the multiplication of two consecutive even numbers.

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\ell $ for some integer ℓ. Now, ${n}^{2}-1=4k(k+1)=8\ell ,$, which proves your result.

To clarify your proof, note that $m=k(k+1)$ must be an even number, so $k(k+1)=2\ell $ for some integer ℓ. Now, ${n}^{2}-1=4k(k+1)=8\ell ,$, which proves your result.

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