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!
 A: 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. 
A: Let $A\equiv \text{$c$ is even}$, $B\equiv \text{$c^4$ is divisible by $16$}$.
The original statement if of the form $\neg B\implies\neg A$ and thus, the contrapositive is $A\implies B$.
However, what you proved is $A\wedge B\implies B$:

By contraposition, let $c$ be an even integer, such that $c=2k$ for
  some integer $k$. [here you assume $A$]
Then, by substitution, 
$(2k)^4 = 16j$, for some integer j, by the definition of divisibility. [here you assume $B$]

(parts in italics added by me)
If you wanted to make this work, you should drop assumption that $j$ is an integer and assume that it is some rational. Later when you write $b = j$, you get that $j$ is not just a rational, but integer number as well, and hence desired result is proven.
A: There are two things wrong with you proof.  (And many things right as well... but two things wrong).
"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."
This is wrong because you don't know that $c^4 = 16j$ for some integer $j$.  This is precisely what you need to prove. It is not a given.
So 
"$(2k)^4 = 16j$".  No, you don't know that any $j$ exist so don't put this in terms of $j$.  Assume nothing and calculate $(2k)^4$.
Like so:
$(2k)^4 =$
$2^4k^4 =$
$16k^4=$
$16b$ for $b = k^4$.
And that's it.  $b = k^4$ is an integer and $c^4 = 16b$
Leave out the $16j$ which you never had in the first place.
"Hence, by divisibility, b=j, which are both integers. "
That's rather meaningless as $j$ never existed.  But also it doesn't matter whether $b = j$ or $b$ is some other integer altogether, just so long as $b$ is an integer.  And $b$ is an integer because k is an integer.  Not because $j$ is. 
If you knew $j$ existed in the beginning you wouldn't need to do any proof at all!  Simply say $(2k)^4 = 16j$ and that's that.  But that's not a proof.  That's simply a statement without justification.
A: The line after "by substitution" is not clear, but it appears to be assuming divisibility, because you say "by divisibility".  There's no need for the $j$ here.  You need to show that $(2k)^4$ is divisible by $16$, so just note that $(2k)^4 = 16k^4$ and $k^4$ is an integer.  So by definition of divisibility, $16$ divides $(2k)^4$ and that proves your contrapositive.
A: 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.
