Proper presentation for proof of $y = x^3 - 3x + 1$ having only irrational roots. Considering the assertion: The polynomial $x^3 - 3x + 1$ has no rational roots, the following is a proof by contradiction:

Let a root of $x^3 - 3x + 1$ be written in the form of $\frac{p}{q}$ where $p$ and $q$ are coprime.
Thus $\frac{p^3}{q^3}$ - $\frac{3p}{q} + 1 = 0$
$\frac{p^3}{q^3}$ - $\frac{3p}{q}  = -1$
By multiplying both sides by $q^3$ we have
$p^3 -3q^2p = -q^3$
$p^3 + q^3 = 3q^2p$
The above equation only holds true if both p and q are even, which I have obtained by trying all 4 cases.
Case 1
p = odd, q = even.
$odd + even \neq 3*odd*even^2 \neq even$
Case 1 is false.
Case 2
p = even, q = odd.
$even + odd \neq 3*even*odd^2 \neq even$
Case 2 is false.
Case 3
p = odd, q = odd.
$odd + odd \neq 3*odd*odd^2 \neq  odd$
Case 3 is false.
Case 4
p = even, q = even.
$even + even = 3*even*even^2 = even$
Case 4 is true.
Now we arrive at the contradiction of the initial condition that $p$ and $q$ are coprime. Thus the polynomial $x^3 - 3x + 1$ has no rational roots.

The question is, what is the proper way to present such a proof?
It seems like writing out all the cases and treating the words odd and even as pseudo variables is a very crude and improper and unofficial way of presenting a proof.
I'm looking for some kind of algebraic presentation that proves $p$ and $q = 2k$ for some $k \in \mathbb Z $
Update
Ok maybe I wasn't very clear in stating what I'm asking for. I'm looking for an algebraic way of proving that $p^3 + q^3 = 3q^2p$ ONLY holds true when both $p$ and $q$ are even numbers, very much unlike the makeshift proof I did above by listing out all 4 cases and checking them one by one.
 A: You can use Rational roots theorem to check that if ${p\over q}$ is a rational root in its lowest form with  $p\in\mathbb Z,\;q\in\mathbb N\,$,
$$
p|1,\;q|1\implies p=\pm1,q=1\implies{p\over q}=\pm1
$$
Plugging in $x=\pm1$ we see $y(1)=-1,y(-1)=3$. Therefore no rational root exists.
A: from here: $p^3 -3q^2p = -q^3$
I would say that the expression on the left is divisible by $p.$  Which means that the expression on the right is divisible by $p.$
Since $p$ and $q$ are co-prime $p = 1.$
A: To undertake your even/odd analysis with a different layout, I would say:

Working $\bmod 2$, 
$$p^3 + q^3 \equiv \begin{cases} 0 & p\equiv q \\
                              1 & p \not\equiv q \end{cases}$$
and
$$3q^2p \equiv \begin{cases} 1 & p\equiv q \equiv 1\\
                              0 & \text{otherwise} \end{cases}$$
Thus only $p\equiv q\equiv0 \bmod 2$ allows $p^3 + q^3 = 3q^2p$, which contradicts the construction that $p$ and $q$ are in lowest terms, as both are divisible by $2$.
A: Example of a different style: 
"Case 1. If $p$ is odd and $q$ is even then $p^3$ is odd and $q^3$ is even, so $p^3+q^3$ is odd but $3pq^2$ is even, so $p^3+q^3\ne 3pq^2.$"
Your  proof is valid. There is nothing illogical about a case-by-case method. 
However we could say $$p^3+q^3=3pq^2\implies p^3+q^3\equiv 3pq^2 \pmod 2\implies$$ $$\implies p+q\equiv pq \pmod 2\implies 1\equiv (p-1)(q-1)\pmod 2\iff$$ $$\implies p-1\not \equiv 0 \not \equiv  q-1\pmod 2 \implies$$ $$\implies p\equiv q\equiv 0\pmod 2.$$
....and therefore $\neg (p\equiv q\equiv 0\pmod 2)\implies p^3+q^3\ne 3pq^2.$
