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

Theorem
Let $p$ be a prime number.

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

Let $k \in \Z_{> 0}: k \ge 2$

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

Proof
Since $x, p > 0$ then $a > 0$.

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

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

Suppose $k$ is odd.

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

Hence $a = \size c^k$

On the other hand, suppose $k$ is even, that is $k = 2l$ for some $l \in Z_{> 0}$.

Then:

In either case $\size c \in \Z_{> 0}$ and $\size c^k = a$

Let $d = \size c$ By the definition of $a$ it follows that $d^k = x^k + p$

Hence:

Let $y = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}$

Since $d, x \in \Z_{> 0}$ then $d - x \in \Z$ and $y \in \Z$

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

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

Since $d, x \in \Z_{> 0}$ then:

Hence $y = p$

Then:

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

Then

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

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