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?
 A: $n^2-1$ must be even $\iff n^2$ must be odd $\iff n$ must be odd
$=2m+1$(say)
Now $(2m+1)^2-1=8\cdot\dfrac{m(m+1)}2\equiv0\pmod8$
Now $n-1=2m\equiv \begin{cases} 0\pmod4 &\mbox{if } m \equiv 0\pmod2 \\ 
2\pmod4 & \mbox{if } m \equiv 1\pmod2 \end{cases}$ 
A: It doesn't have to be complicated! You found a counterexample for $n=7$, so just substitute in:
$$n^2 - 1 = 49 - 1 = 48$$
$$n - 1 = 7 - 1 = 6$$
We know $48$ is divisible by $4$, but $6$ is not divisible by $4$. Therefore, the given statement is false.
A: $n^2 -1$ is a divisible by four is the same as saying $n^2 \equiv 1 \pmod{4}$. Consider the case where $n$ is even. Then $n = 2k$ for some $k \in \mathbb{N}$. Then we have $(2k)^2 = 4k^2 \equiv 0 \pmod{4}$. Hence, we know that $n$ cannot be even. Consider $n$ is odd. Then we have $n = 2k+1$ for some $k \in \mathbb{N}$. Notice $n^2 = (2k+1)^2 = 4k^2+4k+1$. Thus we have $n \equiv 1 \pmod{4}$. So we see that $n$ must be odd.
If $n$ is odd then we have $n = 2k+1$. Notice that $n-1 = 2k$ in this case. So $n-1 \equiv 2k \pmod{4}$. However, there exists $k$ where this is not 0. Notice that $k=1$ we have $n-1 \equiv 2 \pmod{4}$, for $k=3$ we have $n-1 \equiv 2 \pmod{4}$. So this leads us to a claim: if $k$ is odd then this is not true. Assume $k$ is odd. Then we have $k=2j+1$, $j \in \mathbb{N}$. Substitute this in to give us $n-1 \equiv 2(2j+1) \pmod{4}$. However, this gives us $n-1 \equiv 4j + 2 \equiv 2 \pmod{4}$. So if $k$ is odd we have $n-1 \equiv 2 \pmod{4}$. Hence, if $n$ is of the form $2(2j+1)+1 = 4j+3$ where $j \in \mathbb{N}$, this cannot hold true. However, this was too much work; since we found a counterexample, we could've stopped there and said that the claim is not true. Notice that your counterexample is the case where $j=1$. 
A: Let's say,
$$n^2-1=4k$$
$$(n-1)(n+1)=4k$$
Now for that to be true,
Either,

$$(n-1)=4p$$

Or 

$$n+1=4q$$
  Then,
  $$n-1=4q-2\not=4e$$

So , the condition that
$$n^2-1=4n\implies n-1=4q$$
Doesn't hold true.
A: It's simpler with congruences:
The assertion means $n^2\equiv 1\mod 4$ if  and only if $n\equiv 1\mod 4$.
The if part is obviously true. For the only if part, just  list  squares of congruence classes mod. $4$:
$$\begin{array}{l|r r r}
n&0&1&2&3\\\hline n^2&0&1&0&1\end{array}$$
A: Correct. $\,x^2-1\,$ has at least $\color{#c00}2\,$ roots $\bmod m\color{#c00}{>2}\!:\ $ $x\equiv1 $ and $\,x\equiv -1\equiv m\!-\!1.\,$ But if $\,m=2\,$ then $\,-1\equiv 1$ and $\,x^2-1\equiv (x-1)^2$ so $\,x\equiv\,1$ is a double root.
A: 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 $4.$
