$6(6a^2+3b^2+c^2)=5n^2$ has no positive integral solutions

Question

Prove that there doesn't exist positive integers $a,b,c,n$ such that this equality holds:

$6(6a^2+3b^2+c^2)=5n^2$

I found reduced the equation as follows:

$2a^2+b^2+3m^2=10r^2$

But any mod upto $7$ is not useful here.

Answer 1

It is clear $b+m$ is even,

case 1: if $b,m$ are odd numbers,then $$b^2\equiv 1\pmod 8,3m^2\equiv 3\pmod 8,2a^2\equiv 0\textbf{or}2\pmod 8$$ so $$LHS=2a^2+b^2+3m^2\equiv 4\textbf{or} 6\pmod 8$$ but $$RHS=10r^2=0\textbf{or} 2\pmod 8$$ a contradiction.

case 2: if $b,m$ are even number,then let $b=2b',m=2m'$,so $$a^2+2b'^2+6m'^2=5r^2$$ and same as case 1 methos,we have $a,r$must even numbers, so let $a=2a',r=2r'$,then we have $$2a'^2+b'^2+3m'^2=10w'^2$$ but $w'<w$,then use Proof by infinite descent a contradiction.