Find all odd integers $u,v,r,s$ such that $u^2-v^2=3r^2+s^2$

Question

Find all pairwise co-prime odd integers $u,v,r,s$ such that $$u^2-v^2=3r^2+s^2$$

It is easy to confirm that if $u,v,r,s$ are odd, both sides of this equation are a multiple of 4. So it is solvable.

Now, is there an elementary method to find those integers?Any hints will be greatly appreciated.

Answer 1

All solutions of $$ 3 r^2 + s^2 + v^2 = u^2 $$ with $$ \gcd(r,s,v,u) = 1 $$ are given by $$ r = 2 wz + 2 x y, $$ $$ s = w^2 - x^2 + 3 y^2 - 3 z^2, $$ $$ v = 6 y z - 2 wx, $$ $$ u = w^2 + x^2 + 3 y^2 + 3 z^2. $$ If desired, we may switch $s,v.$

For the question asked, this is simply confirming that $r$ and $v$ are even.

Corollary of Theorem 3 in Jones and Pall 1939.

Answer 2

There is no solution.

If $n$ is an odd number then $n^2-1=(n+1)(n-1)$ is a product of two consecutive even numbers. One of these even numbers must be a multiple of $4$ so $n^2-1$ is a multiple of $8$.

Now, if you take modulus $8$: $$u^2-v^2\equiv 1-1=0\pmod 8$$ $$3r^2+s^2\equiv 3+1=4\pmod 8$$