What is wrong with my proof: if $c^4$ is not divisible by 16, then c is odd? Prove that if $c^4$ is not divisible by $16$, then $c$ is odd.

Proof

By contraposition, let $c$ be an even integer, such that $c=2k$ for some integer $k$. 

Then, by substitution, 

$(2k)^4 = 16j$, for some integer j, by the definition of divisibility. 

Then,

$ (2k)^4 = 16j$

$ = 16k^4 = 16j$

$ = 16b = 16j$, where $b$ is an integer and $b=k^4$

Hence, by divisibility, $b=j$, which are both integers. 

Therefore, by contrapositivity, if $c^4$ is not divisible by $16$, then c is odd. 



What is wrong with this proof? This was the answer I wrote down on an exam and only got partial credit. Where am I going wrong? Did I assign too many variables? 

Thanks!
The essence of your argument is correct. You probably lost points for using too many variables. Ideally, a math proof should be as easy to understand as possible. That means using as little notation as possible. 

Here's how I might write the proof:


  Let $c$ be an even integer. Then there exists an integer $k$ such that $c = 2k$. Thus, $c^4 = (2k)^4 = 16k^4$. Since $k$ is an integer, $k^4$ is an integer. Therefore, $c^4$ is divisible by $16$. This completes the proof. 


You probably didn't get full credit on the exam because you didn't explicitly state that $c^4$ is divisible by $16$. That is the most important step of the proof. You proved this (you wrote $(2k)^4 = 16j$), but you should have explicitly written it out.
We can compute this using far fewer variables:

Set $c=2k$, and thus $c$ is even.

Now we can see that $(2k)^4=2^4k^4=16k^4$ which is clearly divisible by $16$

Now we set $c=2j+1$, and thus $c$ is odd. 

Now we can show that $(2j+1)^4 = 16 j^4 + 32 j^3 + 24 j^2 + 8 j + 1$ which we can see is not divisible by $16$

Therefore, if $c$ is not divisible by $16$ then $c$ is odd.