Quadratic Diophantine Equation in Four Variables Consider the equation:  $$d^2 = 6 + a^2 - 3b^2 + 3c^2$$ where $a, b, c, d$ are integers.
Is it necessarily the case that $a$ and $b$ have the same parity and that $c$ and $d$ have the opposite parity to that of $a$ and $b$?  If so, why?
Example 1:  $a = 11, b = 5, c = 12, d = 22$
Example 2:  $a = 14, b = 6, c = 3, d = 11$

Every square is congruent to either $0$ or $1$ $\pmod 4$, hence consider the equation modulo $4$ to get $$\{0,1\} \equiv 2 + \{0,1\} - \{0,3\} + \{0,3\} \pmod{4}$$ where $\{m,n\}$ denotes both possibilities. 
Now if $a$ and $b$ are both even we get $a^2 \equiv 0 \pmod{4}$ and $3b^2 \equiv 0 \pmod{4}$, hence $$\{0,1\} \equiv 2 + \{0,3\} \pmod{4}$$ Easy casework shows that only $d^2 \equiv 1 \pmod{4}$ and $3c^2 \equiv 3 \pmod{4}$ works, so $c$ and $d$ are odd.
If $a$ and $b$ are both odd, then $a^2 \equiv 1 \pmod{4}$ and $3b^2 \equiv 3 \pmod{4}$, so $$\{0,1\} \equiv \{0,3\} \pmod{4}$$ Only possibility is $d^2 \equiv 0 \pmod{4}$ and $3c^2 \equiv 0 \pmod{4}$, i.e. both $c$ and $d$ are even.
