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

Theorem
Let $p \ge 3$ be a prime number.

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

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 $c \in \Z:c^2 = a$.

Since $c^2 \in \Z$, by Nth Root of Integer is Integer or Irrational then:
 * $c \in \Z$

Since $a \neq 0$ then $c \neq 0$

Let $d = \size c$

Then $d \in \Z_{\gt 0}$

From above it follows that $d^2 = c^2 = x^2 + p$

Hence:

Also $d - x_1 \in \Z$ and $d + x \in \Z$

So $d - x$ and $d + x$ are factors of $p$

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

Since $d, x \in \Z_{\gt 0}$ then $d + x \ge 2$

Hence $d + x = p$

It follows that $d - x = 1$

That is, $d = x + 1$

Then

This contrdicts the previous conclusion that $c + x = p$

So:
 * $\nexists \,c \in \Z : c^2 = a$