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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.

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    A not very known fact that might help. If $n\equiv 1\pmod 6$ and $n$ divides $3r^2+s^2$ then $n$ divides $\gcd(r,s)$. (This can be proven by infinite descent, assuming that no such $n$ divides both $r$ and $s$). – 2017-01-07
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    $\gcd(r,s)=1$. So $n=1$, not sure how that can help. Please elaborate. – 2017-01-07
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    I am sorry it is not that obvious but i do mention those integers are all pairwise co-prime. – 2017-01-07
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    It's me who must apologize. I simply did not read the last sentence. – 2017-01-07
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    But thanks for your input. I did not know that fact. – 2017-01-07
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    I just edited the question. – 2017-01-07

2 Answers 2

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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.

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    However that isn't good readable source well. – 2017-01-08
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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 $8$: $$u^2-v^2\equiv 1-1=0\pmod 8$$ $$3r^2+s^2\equiv 3+1=4\pmod 8$$

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    @RossMillikan Indeed, deleting comment now. – 2017-01-07
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    ajotatxe, in case of interest, there is a nice parametrization of all (primitive) solutions, follows fairly quickly from a theorem in a 1939 article. – 2017-01-07