How to prove that the equation $3(a^2+b^2)=7(c^2+d^2)$ has no solution 
Question 1: Let $a,b,c,d$ be positive integers, show that
$$3(a^2+b^2)=7(c^2+d^2)$$ has no solution.

question 1 is from Mathlove, Curious How prove it?

My question 2: Find the least positive integer $k$ such that
$$a^2+b^2=k(c^2+d^2)$$ has no solution.


HINT for Question 1 :
We have $$c^2+d^2\equiv 0\pmod 3\Rightarrow c\equiv d\equiv 0\pmod 3$$ and $$a^2+b^2\equiv 0\pmod 7\Rightarrow a\equiv b\equiv 0\pmod 7.$$
So, since we have $a=7p,b=7q,c=3r,d=3s,$ we have
$$3(a^2+b^2)=7(c^2+d^2)\Rightarrow 7(p^2+q^2)=3(r^2+s^2).$$
