Proof by contradiction: finding integers that satisfy $a+b\sqrt{2}+c\sqrt{3}=\sqrt{6}$.

Question

Assume for contradiction $\exists a,b,c\in\mathbb{Z}$ such that $a+b\sqrt{2}+c\sqrt{3}=\sqrt{6}$.

My current idea for finding a contradiction is to square each side, isolate the square-rooted terms, and repeat. This is pretty cumbersome and therefore I'm wondering if there's an easier way to proving no such integers exist. Any suggestions are appreciated, thank you.

Answer 1

Write it as $b\sqrt{2}+c\sqrt{3}=\sqrt{6}-a $.

Squaring, this becomes $2b^2+2bc\sqrt{6}+3c^2 =6-2a\sqrt{6}+a^2 $ or $(2bc+2a)\sqrt{6} =6-2b^2-3c^2+a^2 $.

Since $\sqrt{6}$ is irrational, this can only hold if $bc+a = 0$ and $6-2b^2-3c^2+a^2 = 0 $.

Since $a = -bc$, the original equation becomes $-bc+b\sqrt{2}+c\sqrt{3}=\sqrt{6} $ or $0 =\sqrt{6}-b\sqrt{2}-c\sqrt{3}+bc =(\sqrt{2}-c)(\sqrt{3}-b) $.

But this implies that either $c = \sqrt{2}$ or $b = \sqrt{3}$, which is impossible.

Answer 2

I think you can first show that if $a+b\sqrt{2}=c+d\sqrt{2}$ where $a,b,c,d\in \mathbb{Z}$, then $a=c,b=d$. Suppose $a+b\sqrt{2}+c\sqrt{3}=\sqrt{6}$, so $(a+b\sqrt{2})^2=3(c-\sqrt{2})^2$. Then you can set up equations relating $a,b,c$ and try to get a contradiction. Since $ab=-3$, there are not many possible cases to try.

Answer 3

The lemma below implies that $\,1,\sqrt2,\sqrt3,\sqrt6\:$ are linearly independendent over $\,\Bbb Q.$

Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\, $ are all $\rm\,\not\in K,\:$ and $\rm\: 2 \ne 0\:$ in $\rm\,K.$

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K,\:$ thus it suffices to show $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K.\:$ But that's impossible, since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ contra, hypotheses, as follows

$\rm\quad\quad\quad\quad\quad\quad\quad\quad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad$ QED