P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2

Theorem
Let $p$ be a prime number.

Let $q \in Z: q \ge 2$

Let $x \in \Z_{\gt 0}: x \ge \dfrac {p + 1} 2$

Let $a = x^q + p$ Then:
 * $a \in \Z_{\gt 0}: \nexists \,c \in \Z : c^q = a$

Proof
Since $x, p \gt 0$ then $a \gt 0$.

for some $c \in \Z:c^q = a$.

Since $c^q \in \Z$, by Nth Root of Integer is Integer or Irrational then:
 * $c \in \Z$

Suppose $q$ is odd.

Since $a \gt 0$, by Odd Power Function is Strictly Increasing then $c \gt 0$

On the other hand, suppose $q$ is even, that is $q = 2n$ for some $n \in Z_{\gt 0}$.

Then:

So it can be assumed that $c \gt 0$

From above it follows that $c^q = x^q + p$

Hence:

Let $y = c^{q - 1} + c^{q - 2} x + c^{q - 3} x^2 + \dotsb + c x^{q - 2} + x^{q - 1}$

Then $c - x \in \Z$ and $y \in \Z$

So $c - x$ and $y$ are factors of $p$

The factors of $p$ by definition are:
 * $\pm 1$ and $\pm p$

Since $c, x \in \Z_{\gt 0}$ then $y \ge 2$

Hence $y = p$

Then:

It also follows that $c - x = 1$, that is, $c = x + 1$

Then

This contradicts the previous conclusion that $p \ge c + x$

So:
 * $\nexists \,c \in \Z : c^q = a$