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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 $4k^2 + 4k$ 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 $4k^2 + 4k + 1 - 1$, which simplifies to $4k^2 + 4k$.

First, we use the distributive property to get the following $4(k^2 + k)$. We let $m = k^2 + k$ and therefore $4(k^2 + k)$ = $4m$. Hence, we know that $4k^2 + 4k$ 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 $4k^2 + 4k$, which means 8 must also be a factor of $4k^2 + 4k$.

$n^2-1$ 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.

bloomsdayforever
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  • $k(k+1)$ must be even numbers. Why? – MH.Lee Nov 05 '21 at 22:13
  • To conclude, you essentially just need to use the fact that if $m$ is even, then $4m$ is divisible by $8$. (And this is quite simple to prove.) – Minus One-Twelfth Nov 05 '21 at 22:14
  • Posting this as a comment, since it has been repeated as an answer many times at mathSE. $(8k+r)^2 \equiv r^2 \pmod{8}$. Therefore, a (valid) alternative approach is to simply manually check the congruences $\pmod{8}$ of $1^2, 3^2, 5^2,$ and $7^2$. – user2661923 Nov 06 '21 at 01:50

3 Answers3

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

Regarding your concern about divisibility, your example is correct and the true theorem is this:

If $a$ and $b$ divide $n$, where $a$ and $b$ are relatively prime, then $ab$ divides $n$.

Sammy Black
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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.

Then ?

dEmigOd
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Your proof is fine. In the number $4k(k+1)$, there are two factors of $2$ in the "$4$," and there is one more factor of $2$ in "$k(k+1)$," so in total there are at least three factors of $2$. The key point is that you've written $4k(k+1)$ as $4 \ \cdot \ k(k+1)$ and checked for factors of $2$ in the factor $4$ and the factor $k(k+1)$ separately.

In your example involving $30$, you've found two factors of the number $30$, which is not the same as what you've done in the proof.

angryavian
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