Prove that $(\sqrt3+2)^m$ is not a natural number for all natural numbers $m≥1$

Question

The aim of this question is to show this lemma:

Prove that $(√3+2)^{m}$ is not a natural number for all natural numbers $m≥1$.

Answer 1

Because $(\sqrt{3}+2)^{m}+(-\sqrt{3}+2)^{m}$ is an integer and $0<(-\sqrt{3}+2)^{m}<1$.

Answer 2

Using the Binomial Theorem, we have:

\begin{align*} \sqrt{3}^m + \binom{m}{1}\sqrt{3}^{m-1}2+\cdots+\binom{m}{m-1}\sqrt{3}2^{m_1}+2^m \end{align*}

The even $m$'s will become rational, but the odd $m$'s will make the $\sqrt{3}$ become a radical, just raised to a higher power. They cannot cancel out because it will always be positive.

Answer 3

I would like to see a comment on if this solution can be improved:

Assume $(\sqrt{3} + 2)^m = p$ so that it is an integer, $n > 0$, a natural number.

$\sqrt{3} + 2 = {p}^{1/m}$

$2 = {p}^{1/m} - \sqrt{3}$

So, in other words, $p$ is a natural number, taken so some power $m \ge 1$, but.

$p^{1/m} - \sqrt{3}$ is irrational, which means, $2$ is irrational. A contradiction.