Find all integer solutions to the Diophantine equation $$a^2+6b^2=3c^2+5d^2$$ Obviously $a=b=c=d=0$ is one solution. For other solutions, I tried reducing mod 4 but ran into too many complicated cases. All I can say is at least one of $a,b,c,d$ must be odd, since if they were all even the factor of $2^2$ can be cancelled out and we can start again.
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If you reduce mod 3 you get $a^2 \equiv 2d^2$, whose unique solution mod 3 is $a \equiv d \equiv 0$. Now, divide by $3$, and continue by infinite descent. – Crostul Dec 17 '16 at 09:13
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http://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation – individ Dec 17 '16 at 11:32
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Try mod $3$ instead.
First note that if any three of $a,b,c,d$ is $0$, then the fourth must be $0$ as well. So if there is a nonzero solution, at least two of $a,b,c,d$ must be nonzero. In this case we can assume WLOG that $a,b,c,d$ have no common factors other than $\pm1$ (otherwise, as you point out, the square of their gcd cancels out).
Now, in mod $3$, the LHS $\equiv0,1\pmod3$ and the RHS $\equiv0,2\pmod3$. Hence, if they're equal they must both be divisible by $3$. It follows that $a$ and $d$ are divisible by $3$, say $a=3a_0$, $d=3d_0$.
Replacing this back and cancelling gives $$3a_0^2+2b^2\ =\ c^2+15d_0^2$$ Again, mod $3$, we have LHS $\equiv0,2\pmod3$ and RHS $\equiv0,1\pmod3$, so $3$ divides both sides, and so $3$ divides $b$ and $c$. Hence $3$ divides $a,b,c,d$, contradicting the fact that $\gcd(a,b,c,d)=1$.
Therefore there are no nonzero solutions; the only solution is $(a,b,c,d)=(0,0,0,0)$.
George Law
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