If $n$ is an integer then $4$ does not divide $n^2 - 3$

Question

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.

Answer 1

For an integer $n$,

$$n^2 \equiv 0,1 \pmod 4$$

and the result follows.

Answer 2

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.

Answer 3

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 $29.5$.

Answer 4

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.

Answer 5

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 $4.$

Answer 6

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.

Answer 7

$$\begin{align} \text{Division Algorithm }\overset{n\,\div\, 2_{\phantom |}}\Longrightarrow n\ \,&= \ r\ +\ 2\, q,\ \ \ \ r\,=\,\color{#0a0}{0,1}\\[0.4em] \text{Squaring the above }\Longrightarrow\, n^2 &= r^2 +\, 4\,Q,\ \ \ r^2 = \color{#0a0}{0,1},\ \ Q\in\Bbb Z\\[.3em] &\equiv\, \color{#0a0}{0,1}\!\!\!\pmod{\!4}\\[.3em] & \not\equiv\, \color{#c00}{3}\\[.3em] \Longrightarrow\ \ 4\nmid n^2\!\!&\ -\color{#c00}3\end{align}\qquad\qquad $$

Remark $\ $ This is a simple special case of Lifting The Exponent.

Answer 8

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 $n=2k-1$ $$\frac{4k^2-4k+1-3}{4}=k^2-k-\frac{1}{2}$$

Answer 9

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.

Answer 10

You need to consider only the following cases:

• $n\equiv\color\red0\pmod4 \implies n^2-3\equiv\color\red0^2-3\equiv-3\equiv1\not\equiv0\pmod4$
• $n\equiv\color\red1\pmod4 \implies n^2-3\equiv\color\red1^2-3\equiv-2\equiv2\not\equiv0\pmod4$
• $n\equiv\color\red2\pmod4 \implies n^2-3\equiv\color\red2^2-3\equiv+1\equiv1\not\equiv0\pmod4$
• $n\equiv\color\red3\pmod4 \implies n^2-3\equiv\color\red3^2-3\equiv+6\equiv2\not\equiv0\pmod4$

Answer 11

"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.

Answer 12

We have to show that

$\tag 1 n \ge 0 \text{ implies } 4\nmid n^2 -3$

This can be accomplished using Fermat's method of descent.

We check the first two cases:

$\; \text{For }n = 0,\, n^2 -3 = -3 \text{ and } 4\nmid -3$

$\; \text{For }n = 1,\, n^2 -3 = -2 \text{ and } 4\nmid -2$

Suppose that for $4\mid n_0^2 -3$ for some $n_0 \gt 1$. Since

$\; (n_0^2 - 3) - ((n_0 - 2)^2 - 3) = 4 (n_0 - 1)$

it would have to also be true that $4\mid (n_0-2)^2 -3$ and $n_0 - 2 \gt 1$.

This infinite descent results in a contradiction.