If $n$ is an integer then $4$ does not divide $n^2 - 3$ Having some trouble proving this. Was going to attack with using the contrapositive of this statement but can't seem to show that $n$ isn't an integer. 
 A: For an integer $n$, 
$$n^2 \equiv 0,1 \pmod 4$$
and the result follows.
A: We can prove this by contradiction.
Assume that $4$ divides $n^2 -3$
$$4|n^2-3,$$ then $$4m=n^2-3,$$ where m is some integer. Now let's break the this into two cases.
Case n is even (k is an integer):
$$n = 2k$$
$$(2k)^2-3=4m$$
$$4k^2-3=4m$$
$$k^2-3/4=m$$
we know that the integers are closed under multiplication, so $k^2$ is an integer but clearly since $3/4$ is rational and not an integer $k^2-3/4$ is also not an integer. Which contradicts 'm is an integer'.
Case n is odd (k is an integer):
$$n = 2k +1 $$
$$(2k+1)^2-3=4m$$
$$4k^2+4k+1-3=4m$$
$$2k^2+2k-1=2m$$
$$k^2+k-1/2=m$$
we know that the integers are closed under multiplication and addition, so $k^2+k$ is an integer but clearly since $1/2$ is rational and not an integer $k^2+k-1/2$ is also not an integer. Which contradicts 'm is an integer'.
Okay now we have proved that $4$ doesn't divide $n^2-3$, for n equal to odd or even, which would be all the integers.
A: You just need to look at two cases: $n$ is odd and $n$ is even.


*

*If $n$ is even, then $n \equiv 0 \textrm{ or } 2 \pmod 4$ but $n^2 \equiv 0 \pmod 4$ regardless, which means that $n^2 - 3 \equiv 1 \pmod 4$, and therefore $$\frac{n^2 - 3}{4} = m + \frac{1}{4},$$ where $m$ is some integer. For example, $n = 10$ gives us $24.25$.

*If $n$ is odd, then $n \equiv 1 \textrm{ or } 3 \pmod 4$ but $n^2 \equiv 1 \pmod 4$ regardless, which means that $n^2 - 3 \equiv 2 \pmod 4$, and therefore $$\frac{n^2 - 3}{4} = m + \frac{1}{2},$$ where $m$ is some integer. For example, $n = 11$ gives us $29.5$.

A: I think the easiest way is by simple algebraic manipulation. Given some integer $k$, if $n = 2k$ (meaning that $n$ is even), then $n^2 - 3 = 4k^2 - 3$; if $n = 2k + 1$ (meaning $n$ is odd), then $n^2 - 3 = 4k^2 + 4k - 2$. Notice that $4k^2$ and $4k^2 + 4k$ are both multiples of 4, while $4k^2 - 3$ and $4k^2 + 4k - 2$ are not.
A: case 1:  $n$ is even
$n^2$ is divisible by $4.\: n^2 - 3$ is not divisible by $4.$
case 2: $n$ is odd
$n^2 - 3 = (n^2 - 1) - 2 = (n+1)(n-1) - 2$
If $n$ is odd $(n+1)(n-1)$ is divisible by $4$ and $(n+1)(n-1) - 2$ cannot be divisible by $4.$
A: If $n$ is even, then $n^2$ is a multiple of 4, hence $n^2 - 3$ is not.
If $n$ is odd, then write $$M = n^2 - 3 = [(n-1)(n+1)] - 2.$$  The product in []'s is a product of two even numbers, hence is a multiple of 4.  Therefore, $M$ will give a remainder of 2 when divided by 4.
A: $$\begin{align} \text{Division Algorithm }\Rightarrow\ n\ \,&= \ r\ +\ 2\ q,\quad\ \ \ r\,=\,\color{#0a0}{0,1}\\[0.4em] \text{Squaring the above }\Rightarrow\, n^2 &= r^2 +\, 4(\cdots),\ \ r^2 = \color{#0a0}{0,1}\\[.3em]
&\,\equiv\, \color{#0a0}{0,1}\!\!\!\pmod{\!4}\\[.3em]
&\ \not\equiv\, \color{#c00}{3}\\[.3em]
\iff 4&\nmid n^2-\color{#c00}3\end{align}\qquad  $$
A: Hint:
For even numbers
let $n=2k$
Where $k$ integer number takes $1,2,3,...$
$$\frac{4k^2-3}{4}=k^2-\frac{3}{4}$$
For odd numbers
let $n=2k-1$
$$\frac{4k^2-4k+1-3}{4}=k^2-k-\frac{1}{2}$$
A: Noting that$$(n\pm 2)^2=n^2\pm 4n+4\equiv n^2 \;({\text{mod }}4),$$
the statement holds for $n\pm 2$ if it holds for $n$, and so if it ever fails it must fail for either $n=1$ or $n=2$.  But $1^2-3=-2$ and $2^2-3=1$, neither of which is divisible by $4$; so the statement always holds.
A: You need to consider only the following cases:


*

*$n\equiv\color\red0\pmod4 \implies n^2-3\equiv\color\red0^2-3\equiv-3\equiv1\not\equiv0\pmod4$

*$n\equiv\color\red1\pmod4 \implies n^2-3\equiv\color\red1^2-3\equiv-2\equiv2\not\equiv0\pmod4$

*$n\equiv\color\red2\pmod4 \implies n^2-3\equiv\color\red2^2-3\equiv+1\equiv1\not\equiv0\pmod4$

*$n\equiv\color\red3\pmod4 \implies n^2-3\equiv\color\red3^2-3\equiv+6\equiv2\not\equiv0\pmod4$

A: "Was going to attack with using the contrapositive of this statement but can't seem to show that n isn't an integer. "
That's a weird way of doing it but:
If $4|n^2 - 3$ then $4|n^2 + 1$ so $n^2/4 + 1/4 = k $ for some integer $k$ so $n^2 = 4k -1$. $n^2$ must be odd. If $n$ is integer then $n$ is odd so $n= 2m + 1$ for some integer $m$.  So $(2m + 1)^2 = 4k - 1$ so $ 4m^2 + 4m + 1 = 4k - 1$ so $4m^2 + 4m = 4k - 2$ so $2m^2 + 2m = 2k -1$.  The left hand side is even and the right hand side is odd.  A contradiction.
So $n$ is not an integer. 
