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 $a > 0$.
Aiming for a contradiction, suppose 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:
\(\ds a\) | \(=\) | \(\ds c^{2l}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {c^2}^l\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\size c^2}^l\) | Equivalence of Definitions of Absolute Value Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \size c^{2l}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size c^k\) |
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:
\(\ds p\) | \(=\) | \(\ds d^k - x^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {d - x} \paren {d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + c x^{k - 2} + x^{k - 1} }\) | Difference of Two Powers |
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:
\(\ds d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}\) | \(\ge\) | \(\ds d^{k - 1} + x^{k - 1}\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds d + x\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds 1 + 1\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds 2\) |
Hence $y = p$
Then:
\(\ds p\) | \(=\) | \(\ds d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds d^{k - 1} + x^{k - 1}\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds d + x\) |
It also follows that $d - x = 1$, that is, $d = x + 1$
Then
\(\ds d + x\) | \(=\) | \(\ds \paren {x + 1} + x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2x + 1\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 2x\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds p + 1\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds p\) |
This contradicts the previous conclusion that $p \ge d + x$
So:
- $\nexists \,c \in \Z : c^k = a$
$\blacksquare$