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.

 A: 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
$$3(a^2+b^2)=7(c^2+d^2)\Rightarrow 7(p^2+q^2)=3(r^2+s^2).$$
A: $$\frac{a^2+b^2}{c^2+d^2}=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}$$
Use Brahmagupta–Fibonacci identity for the numerator to get $$\frac{a^2+b^2}{c^2+d^2}=\left(\frac{ac\pm bd}{c^2+d^2}\right)^2+\left(\frac{ad\mp bc}{c^2+d^2}\right)^2$$
But $3,7\equiv-1\pmod4$  can not be expressed as the sum of two squares 
A: The famous two-square theorem by Fermat implies that $3n$ is not the sum of two squares, if $n$ is. This is answers Question $2$.
The first question is a nice example of infinite descent. One can show that all four numbers are dibisible by $3$, if the equation holds. Hence we get a smaller solution by dividing the equation by $9$.
A: Q1: c and d are both $0\mod3$, so RHS is divisible by 9, which means a and b are both $0 \mod 3$, so LHS divisible by 27, etc...
Q2: k=3 by similar argument to Q1.
