Proof problem: Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.

Since $n$ is odd $n=4m+1$ or $n=4m+3$.

In the first case $n^2-1=(n-1)(n+1)=4m\cdot(4m+2)=8m(2m+1)$, while in the second case $n^2-1=(n-1)(n+1)=(4m+2)\cdot(4m+4)=8(2m+1)(m+1)$.

So $n^2-1$ is divisible by 8 if $n$ is odd.

If $n$ is an odd positive integer, then $n=2m+1$ for some non-negative integer $m$. Can you see how to finish? Hint- Do two cases , one where $m$ is odd and one where $m$ is even.

This is a nice example of a direct proof. You start with the facts that if $\phi$ is your positive odd integer, then it is in the form $\phi = 2n + 1$ where $n$ is an integer and $\phi^2 - 1 = 8p \quad( p\in\mathbb Z) $ is true.

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Recall that if a number is divisible by 8, then 8 is one of its factors.

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This is something like a “case-by-case” proof. You prove the hypothesis for $n \in2\mathbb Z$ and then $n \in 2\mathbb Z + 1$.

Let's start with assuming that $n = 2m + 1$ where $m$ is an integer (of course) to prove for odd numbers. It is clear that,$$\begin{align} \phi^2 - 1 =(2n + 1)^2 - 1 &= 4n^2 + 4n \\ &= 4n(n + 1) \\ &= (8mn + 4n)(2m + 2) \\ &=4n\cdot 2\cdot(2m + 1)\cdot (m + 1) \\ & = 8 \cdot (n(2m + 1)(m +1)) \\ & = 8k \end{align} $$ $\rm 0.5\times Q.E.D $

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Now $n = 2m$ which is gonna be easier. $$\begin{align}\phi^2 - 1 =(2n + 1)^2- 1 = 4n^2 + 4n &= 4n(n + 1) \\ &= 8mn(2m +1)\\&= 8(2m^2n + mn) \\ &=8k \end{align}$$

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Q.E.D

(yay)