Proof verification: $\forall n \in \mathbb{Z}, 4\nmid(n^2+2)$

Question

Would someone be willing to verify this proof by contradiction? NB: This theorem has appeared on MSE before, but my proof differs.

Theorem: $\forall n \in \mathbb{Z}, 4\nmid(n^2+2)$

Suppose, for the sake of contradiction, that the above theorem is false. Then, $\exists n \in \mathbb{Z}, 4|(n^2+2)$.

$4|(n^2+2) \Longrightarrow 4x=(n^2+2), x \in \mathbb{Z}$.

It follows that $4x-2 = n^2 \Longrightarrow \sqrt{2}\sqrt{2x-1} = n$. $\sqrt{2}$ is irrational, so for the product $\sqrt{2}\sqrt{2x-1}$ to be an integer, the term $\sqrt{2x-1}$ must contain a $\sqrt{2}$ as one of its factors. Edit to add for clarity: In other words, for a product $\sqrt{2}a$ to be integer, we need to be able to state $a = \sqrt{2}b$, so $\sqrt{2}\sqrt{2}b \Longrightarrow 2b$. (Or does this argument not hold water?)

However, $2x-1$ is an odd number by definition, and therefore cannot have 2 as a factor. Therefore, $n \notin \mathbb{Z}$, and we have a contradiction.

Answer 1

The $\sqrt 2$ as a factor argument needs help. Try using cases.

If $n$ is even then $n = 2k$ for some $k$ and $n^2 + 2 = 4k^2 + 2$.

If $n$ is odd then $n = 2k+1$ for some $k$ and $n^2 + 2 = 4k^2 + 4k + 3$.

I don't think either of these is a multiple of $4$.

Answer 2

I don't think this is a good proof; in particular the part where you state "for the product $\sqrt{2}\sqrt{2x-1}$ to be an integer, the term $\sqrt{2x-1}$ must contain a $\sqrt{2}$ as one of its factors" is not convincing. What is a factor of an irrational number?

I would suggest one of two alternate approaches.

$\qquad (1)$: Try induction.

$\qquad (2)$: Consider odd $n$ and even $n$ separately.

Answer 3

Note that either

$$n^2\equiv 0 (\mod 4)$$ Or

$$n^2\equiv 1 (\mod 4)$$ Giving $$n^2 +2\equiv 2 (\mod 4)$$ Or

$$n^2+2\equiv 3 (\mod 4)$$

Answer 4

You have made it somewhat complicated while there is a simpler solution.

$$\forall n\in\mathbb{Z}; 4\nmid(n^2+2)$$

Assume that: $\exists n\in\mathbb{Z}; 4\mid(n^2+2)$. So we can write ($k\in\mathbb{Z}$):

$$n^2+2 = 4k\implies n^2 = 4k - 2\implies n^2 = 2(2k - 1)\implies 2\mid n^2\implies 2\mid n$$

So $n = 2m$ ($m\in\mathbb{Z}$). Therefore:

$$(2m)^2 + 2 = 4k\implies 4m^2 + 2 = 4k\implies 4m^2 = 4k - 2\implies 2m^2 = 2k - 1\implies 2k - 2m^2 = 1\implies \color{red}{2(k - m^2) = 1}$$

This implies that $2\mid 1$. So our first assumption is false and hence the statement is proved. $\blacksquare$

Answer 5

My proof by contradiction,

Proof: Suppose $4|(n^2+2)$, there are two cases as $n$ is either even or odd.

Case 1: $n$ is even, $n=2k$ $$4|((2k)^2+2)=4k^2+2 $$ $$ 4 \nmid 2 \Rightarrow 4 \nmid 4k^2+2$$ This is a contradiction.

Case 2: $n$ is odd, $n=2k+1$ $$4|((2k+1)^2+2)=4k^2+4k+1+2=4k^2+4k+3 $$ $$ 4 \nmid 3 \Rightarrow 4 \nmid 4k^2+4k+3$$ This is a second contradiction.

As there is a contradiction in both cases this implies that $ 4\nmid(n^2+2), \ \forall \ n \in \mathbb Z$