How to prove a square number minus $2$ will never divide $4$ evenly

Question

For every $n$ that is an element of the set of all integers, prove that $4$ will never divide $n^2-2$.

Not sure how to add in the characters but basically I am looking for a proof that $\dfrac {n^2-2}4$ will never equal an integer.

I tried and don't really know how to turn this into a proof for every integer

What I tried to do was follow a previous example I did where a|x= ka so I put $(n^2)=k$ and $(-2)=l$ so $4\mid (n^2-2)= 4\mid (k^2-l)$ but I realized this was getting me nowhere so I came to ask on here

Answer 1

For a number to divide by $4$ evenly, it must be even.

So we disproved the statement for all $n=\text{odd}$, since $\dfrac{\text{odd}^2-2}{4}=\text{not whole}$.

For $n=\text{even}$:

$$\dfrac {n^2-2}4=\dfrac{n^2}{4}-\dfrac 12=\left(\dfrac{n}{2}\right)^2-\dfrac 12$$

Since $\left(\dfrac{n}{2}\right)^2$ is a whole number for all $n=\text{even}$, what must $\left(\dfrac{n}{2}\right)^2-\dfrac 12$ be?

Answer 2

HINT

Case 1. $n$ is odd. Then $n^2$ is odd...

Case 2. $n$ is even, then $n=2k$ so $$\frac{n^2-2}{4} = \frac{4k^2-2}{4} = \frac{2k^2-1}{2}$$

Answer 3

Hint: $$n=4k+i,$$ where $i=0,1,2,3$, and $$(4k+i)^2 -2 = 16k^2+8k+i^2-2 =4(4k^2+2k) + (i^2-2).$$

Answer 4

In terms of congruences, the statement $4 \mid n^2 - 2$ means that $n^2 \equiv 2 \bmod 4$. Check for $n = 0, 1, 2,$ and $3$ that $n^2 \not\equiv 2 \bmod 4$.