There exists an integer k such that $n = 3k+1$. Then $n^2 = (3k+1)^2 =9k^2 + 6k + 1 = 3 (3k2 +2k)+1$. Consider the following proof fragment.

There exists an integer $k$ such that $n = 3k+1$. 

Then $n^2 = (3k+1)^2 =9k^2 + 6k + 1 = 3 (3k^2 +2k)+1$.

For each of the statements, $(a), (b), (c)$, below, answer the following. 

Does the fragment provide a proof of the statement? If yes, explain why. If no, explain why not. The letter n denotes an integer.

(a) If $n$ is odd, then $n^2$ is odd.

(b) If $n^2$ is divisible by $3$, then $n$ is divisible by $3$.

(c) If $n$ leaves remainder $1$ on division by $3$, then so does $n^2$.

I am having some trouble understanding the question.
I have started a) by assuming $n$ is odd and I took $n=3k+1$ and then I showed that $n^2=(3k+1)^2=3(3k^2+2k)+1$ 
and I said that if $n=3k+1$ is odd $3k$ is even so $k$ is even and can be writen as $k=2m$ where $m$ is an integer. After that I sub. $k=2m$ into the $n^2=3(3k^2+2k)+1$ and I find that $n^2=2$(somthing)$+1$ so it is odd.

I have been told by my Professor that is I should not use the statment given to me to answer $a,b$, and $c$. I was wondering if for $a)$ I am just supposed to take $n=2k+1$ and then show that $n^2$ is odd.
Can someone help me out?
Generally, what does it mean for a number to be odd? Well mathematically we say:

$(2k)$ is even, and $(2k+1)$ is odd, for $k\in \mathbb{N}$.

Let's consider part (a):

Prove if $n$ is odd, $n^2$ must also be odd:

Let n be an odd number, such that $n=2k+1$ for $k\in \mathbb{N}$.

Then, $n^2=(2k+1)^2=(4k^2+4k+1)=2(2k^2+2k)+1\Rightarrow n^2$ is odd.

Why? Well any natural number multiplied by an even number is even, so what happens when we add 1 to an even number? It turns odd!

Consider part (b):

Prove if $n^2$ is divisible by $3$, then $n$ is divisible by $3$.

If $n^2$ is divisible by $3\Rightarrow n^2=3k$ for $k\in \mathbb{Z}$ 

Well, this comes directly from the fundamental theorem of arithmetic, that is, every number is either prime, or the product of prime numbers, such that each composite number has a unique sequence of products. (excluding the order in which primes are multiplied of course).

Well, this implies that $n^2$ is the product of a sequence of prime numbers, that is, $n^2$ is a number multiplied by itself. So intuitively it makes sense that $n$ must also be divisible by $3$, and that's the proof!  

Consider part (c):

Can you express $n^2=3M+R$ such that $M\in \mathbb{N}$ and $R$ is your remainder, in this case $R=1$?
Question has been answered in the comments by JMoravitz.