Prove there exists no rational solutions $a,b,c$, and $d$ such that $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$

Question

I want to prove there exists no rational solutions $a,b,c$, and $d$ such that $$a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$$I have proven there exists no rational solutions $a,b$, and $c$ such that $$a+b\sqrt2=\sqrt3$$ $$a + b\sqrt2 + c\sqrt3 = \sqrt6$$ $$a + b\sqrt2 + c\sqrt3 = \sqrt5$$ A hint would be much appreciated.

Answer 1

I have proven there exists no rational solutions $a,b$, and $c$ such that [...] $a + b\sqrt2 + c\sqrt3 = \sqrt6$

Hint: if you square $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$ then you'll get an equality of precisely that form.

Answer 2

$(b\sqrt{2}+c\sqrt{3})^2 = (\sqrt{5} - a - d\sqrt{6})^2\implies 2b^2+2bc\sqrt{6}+3c^2= 5+a^2+6d^2 - 2a\sqrt{5}-2d\sqrt{30}+2ad\sqrt{6}\implies m+r\sqrt{6}= q\sqrt{5}+t\sqrt{30}\implies (r\sqrt{6} - q\sqrt{5})^2 = (t\sqrt{30}-m)^2\implies 6r^2-2qr\sqrt{30}+5q^2= 30t^2-2tm\sqrt{30}+m^2\implies \sqrt{30} \in \mathbb{Q}$, a contradiction.

Answer 3

It is enough to exploit the existence of infinite primes $p$ such that both $2$ and $3$ are quadratic residues $\!\!\pmod{p}$ while $5$ is not. By the Chinese remainder theorem and quadratic reciprocity, there are infinite primes of the form $p=120k+47$ and anyone of them does the job. Assuming $$ a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=\sqrt{5},\qquad a,b,c,d\in\mathbb{Q}$$ we have that as soon as $p$ is large enough and both $x^2-2$ and $x^2-3$ completely factors over $\mathbb{F}_p$, $x^2-5$ does the same, leading to a contradiction.