Decide if the statement " $n^2-1$ is multiple of $4$ if and only if $n-1$ is multiple of $4$" is true or false. The statement is biconditional ( P $\Longleftrightarrow$ Q )
$n-1$ is multiple of 4 $\Longleftrightarrow n^2-1$ is multiple of 4
The statement is true if $\,$  P $\Rightarrow$ Q and  Q$\Rightarrow$ P are true

P $\Rightarrow$ Q
$ n-1=4k, \,  k  \in \mathbb{N} \Rightarrow n^2-1=4p\, , p   \in \mathbb{N}  $
Direct Proof:
\begin{split}
 n-1=4k, \, k  \in \mathbb{N} \Rightarrow n^2-1  & = (n-1)(n+1)\\
 & = 4k\, (n+1) \\ & =4kn+4k \\ &=4(kn+k) \\&=4p, \, p \in \mathbb{N}
\end{split} 
$\therefore \,$  P $\Rightarrow$ Q is true

Q $\Rightarrow$ P
$ n^2-1=4p, \,  p  \in \mathbb{N} \Rightarrow \, n-1=4k\  , k   \in \mathbb{N}  $
I found  a counterexample for $n=7$ 
 but I don't know how to proove it with a formal method.
Can someone help me please?

1)$n^2-1= (n-1)(n+1)$, 
Let $n$ be odd: $n+1, n-1$ are even:
$n+1= 2r$, $n-1= 2s$.
$n^2-1= 4rs$ is divisible by $ 4$.
Choose $n=7:$ 
$n-1 = 6$,  is not divisible by $4$.
2) let $n-1$ be divisible by $4$, I.e. $n-1 =4k$.
Since $n^2-1=(n-1)(n+1)=$
$4k(n+1)$, hence
$n^2-1 $  divisible by $4$.
Recapping: 
False:
1) $n^2-1$ is divisible $ 4$ $ \Rightarrow$  $n-1$ is divisible by $4$.
(counterexample: $n=7$).
True:
2) $n-1$ is divisible by $4$ $\Rightarrow$ $n^2 -1$ is divisible by $4.$
