Proof: $^2 + 2$ isn't divisible by 4 for all integers n I assumed by contradiction that $n^2 + 2 = 4k$, where k is any integer.
$n^2 = 4k - 2$, so $n^2 = 2(2k-1)$. This means $n^2$ is even, so n is even, but we assumed this is true for all integers n, but it isn't true for odd integers, which is a contradiction.
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Is this correct ?
 A: No, this is a mixing up of "all" and "some". You have proven that it cannot be true that for every integer $n$, $n^2-2$ is divisible by 4. (In fact you've shown it's not true if $n$ is odd, because then $n^2-2$ is odd.) But the question is about proving that it is not true that for any integer $n$, $n^2-2$ is divisible by 4.
That is, the claim is wrong if you can find even one value of $n$ such that $n^2-2$ is divisible by $4$. And to prove the claim true, you need to make an argument about every single value of $n$.
You've made a good argument for odd $n$, so now consider even $n$. Why can't $n^2-2$ be a multiple of $4$ if $n$ is even? (Try some small examples like $n=2$, $n=4$, $n=6$ if you get stuck.)
A: No.  You didn't assume it was true for all integers.  In order to prove something is true for all integers, and use contradiction, you assume it's false for one integer.  Then you show that even for this one integer, it doesn't work.
You correctly conclude that $n$ must be even.  That means $n=2m$.  Plug that in and see if you can get your contradiction.
A: The problem in your reasoning has been explained well by other users, so I will complement the other answers by providing two alternative solutions: one direct, and one by contradiction
Alternative solution :
Lemma: any perfect square leaves a remainder of either $0 \ \text{or}\  1$ on division by $4$. (If you know modular arithmetic, you may recognise it to be the fact that any perfect square $a^2$ satisfies $a\equiv 0\ \text{or}\ 1\bmod 4$ where $a\in \mathbb Z_+$).
Proof of lemma: If $a$ is odd then we can write $a=2n$ where $n$ is a positive integer. Then, $a^2=4n^2$ which is completely divisible by $4$ and thus leaves the remainder of $0$ upon dividing by $4$. If $a$ is odd, thn $a=2n+1$ and so $a^2=4(n^2+n)+\fbox{$\color{red}1$}$ and so always leaves a remainder of $1$ upon division by $4$. Since the union of the sets of odd numbers and the sets of even numbers give us the set of positive integers, hence proved.
By the lemma, $a^2+2$ leaves a remainder of either $1\ \text{or}\ 2$ when divided by $4$ ( s $a$ is even or odd respectively) so is never divisible by $4$. $\blacksquare$
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PROOF BY CONTRADICTION (alternative):
Suppose there exists such a positive integer $m$ for which $m^2+2$ has the audacity to be fully divisible by $4$. Now, a number divisible by 4 must necessarily be even, and therefore $m^2+2$ is even.
If $m^2+2$ is even, then $m^2$ must also be even (make sure you can rigorously explain why). This implies that $2|m^{\color{magenta}2}$.
Thus, we have, $2|m$. (For a proper reasoning as to why, refer to this page: How to prove if $n$ is prime and $n | a^2$ then $n | a$?) Thus $m$ is even and can be written as $m=2n$ where $n$ is another positive integer. If so, then $m^2=4n^2$ is divisible by $4$ and hence, $m^2+2$ always leaves a remainder of $2$ when divided by $4$.
This implies that $m^2+2$ is not divisible by $4$, which is a contradiction. Thus no such $m$ exists. $\blacksquare$
