<s>[INST] Prove there exists no rational solutions $a,b,c$, and $d$ such that $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$ 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.
 [/INST] $(b\sqrt{2}+c\sqrt{3})^2 = (\sqrt{5} - a - d\sqrt{6})^2\implies 2b^2+2bc\sqrt{6}+3c^2= 5+a^2+6d^2 - 2a\sqrt{5}-2d\sqrt{30}+2ad\sqrt{6}\implies m+r\sqrt{6}=  q\sqrt{5}+t\sqrt{30}\implies (r\sqrt{6} - q\sqrt{5})^2 = (t\sqrt{30}-m)^2\implies 6r^2-2qr\sqrt{30}+5q^2= 30t^2-2tm\sqrt{30}+m^2\implies \sqrt{30} \in \mathbb{Q}$, a contradiction.
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