daktielti

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, $\left(2k+1{\right)}^{2}-1$ must be divisible by 8 where $n=2k+1$ for some integer k. Next, we expand $\left(2k+1{\right)}^{2}-1$ 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 $4\left({k}^{2}+k\right)$. We let $m={k}^{2}+k$ and therefore $4\left({k}^{2}+k\right)$. 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 $4k\left(k+1\right)$. 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.

Ashley Parks

Expert

Explanation:
Basically, you only need to know the following fact:
${n}^{2}-1=\left(n-1\right)\cdot \left(n+1\right).$
And above is the multiplication of two consecutive even numbers.

Augustus Acevedo

Expert

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

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