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

The aim of this question is to show this lemma:

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

• Hint: Use the Binomial theorem. – Hakim Oct 25 '14 at 17:12
• You can show that $\{a + b\sqrt{3}, a, b \in \mathbb{N}^*\}$ is closed under multiplication – Petite Etincelle Oct 25 '14 at 18:31
• – punctured dusk Apr 7 '15 at 12:41

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

• Note that this also shows that the expression becomes arbitrarily close to an integer. – Mark Bennet Oct 25 '14 at 17:32
• and compute the fractional part of this number... – Chen Jiang Oct 25 '14 at 17:44
• @ChenJiang: can you please elaborate about this last comment. – DER Nov 5 '14 at 17:00
• a+b is an integer and 0<b<1 implies that the fractional part of a is just 1-b – Chen Jiang Nov 6 '14 at 3:06

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.

• This isn't enough, there's no reason to think that the odd powers of $\sqrt 3$ won't add up to an integer. – Git Gud Oct 25 '14 at 17:18
• @GitGud Assume $k$ is odd then $\sqrt{3}^k=\sqrt{3}^{2n+1}=3^n\cdot\sqrt3$. So those odd parts can be written as $q\sqrt3$ for some whole number $q$, and factoring will yield $\sqrt{3}\cdot\text{a whole number}$ which is irrational. – Hakim Oct 25 '14 at 17:21
• @Hakim I had missed that, but I stand by my comment, what the answer has written isn't enough. – Git Gud Oct 25 '14 at 17:23
• @GitGud Could you clarify why please? Maybe I also missed a point. – Hakim Oct 25 '14 at 17:31
• @Hakim I mean that it's not sufficiently justified, specially considering the answer mentions the non-negativity of numbers which implies the impossibility of cancelation. – Git Gud Oct 25 '14 at 17:34

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.

• the last sentence is not clear. "irrational" part – Chen Jiang Oct 25 '14 at 17:30
• @ChenJiang. rational plus irrational is irrational. And if $p^{1/m}$ is irrational! the sum will still be irrational On the LHS, I had $2$ concluding that the RHS is irrational would conclude that the LHS $=2$ is irrational as well! which is a false contradiction. – Amad27 Oct 25 '14 at 18:21
• "And if $p^{1/m}$ is irrational! the sum will still be irrational" this sentence is not so clear. – Chen Jiang Oct 25 '14 at 18:23
• @ChenJiang,sorry, I'll try to be clear here. If $p^{1/m}$ is irrational, provided that $p^{1/m} \ne \sqrt{3}$ the sum, $p^{1/m} - \sqrt{3} = x$ will prove to be that $x$ is in irrational number, which is of course, not a natural number. This concludes the RHS is irrational. But the LHS = $2$, which means that it says if RHS is irrational then LHS is irrational, which is false because $2$ is rational, a natural number. – Amad27 Oct 25 '14 at 18:36