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.
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.
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
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.
