If $n$ is such a positive integer, that $8|n^2$, then $4|n$ I'm new to the subject of discrete mathematics.
This statement is either true or false and it has to be proved. I've struggled with this exercise for quite a while, and this is what I came up with: 


*

*If $8|n^2$ then $n^2$ is even

*If $n^2$ is even then $n$ is even

*If $n$ is positive and $4|n$ then $n = 4k$ ($k$ - any positive integer)

*If $n = 4k$ then $n^2 = 16k^2$

*$8|16k^2$ and $4|4k$, therefore the statement is true


$n|m$ means $n$ divides $m$
Can someone verify whether I proved it or not?
 A: If $k$ is the number of factors $2$ that $n$ has in its prime factorisation, then we know that $n^2$ has $2k$ such factors.
We are given that $8$ divides $n^2$ so $2k \ge 3$. As  $k$ is an integer this means that $k \ge 2$, so indeed $4|n$.
A: No, you did not prove it. At no point what you wrote as something equivalent to “… and therefore $4\mid n^2$”.
You stated (correctly) that $n$ is even. How could $n$ then fail to be a multiple of $4$? Only if $n=2k$, whre $k$ is an odd number. But then $n^2=4k^2$ and, since $k^2$ is odd too, $8\nmid4k^2$. Thereby, a contradiction is reached and so $4\mid n$.
A: Sorry, but your proof is wrong, because at a certain point you assume that $4\mid n$.
Since $8\mid n^2$, $n$ is even, so $n=2a$. Hence $8\mid 4a^2$, which implies $2\mid a^2$. Therefore $a$ is even: $a=2b$ and finally $n=4b$.
A: Hint: the number of primes in the prime factorization of a square is even.  Now $8=2^3$.
A: No, you have not prooved it.
What you've prooved is the reverse implication (if $4|n$, then $8|n^2$) and two first steps of your reasoning seems to be not used at all.
There are two hints, that may help you start the proper proof (two alternative proofs - try to do them both). I think you can manage how to continue the proof from these points.
Hint 
You can start your proof by noticing, that $n=m2^k$ for some integer $k$ and odd $m$.
Hint
Alternatively you can start your proof by noticing, that if $a=b\cdot c$ and $8|a$, then $4|b$ or $4|c$.
