a long time ago I posted a question to find a general solution to a modified Pythagorean equation, mainly $a^2+b^2=2c^2$ that question was eventually answered. But now I need more help.
I now have 3 separate equations, and I would like a general solution for each (independently for each single version).
$$2a^2+b^2-c^2=2d^2 \tag {eq.1}$$ $$2a^2+b^2+c^2=4d^2 \tag {eq.2}$$ $$3a^2+2b^2-c^2=4d^2 \tag {eq.3}$$
Any help would be appreciated since I have no idea how I would come up with general solutions to these. Thank you in advance.
Also, I will only consider any solution where all numbers $a,b,c,d$ in all equations are pairwise different (and all are not zero).
PS. for those asking about the tags, yes this is related to 3x3 magic squares.
The first equation can be rearranged to $$2a^2-2d^2 = c^2 - b^2\\2(a-d)(a+d) = (c-b)(c+b)$$
If $d > a$, then we can exchange the values of $a$ and $d$, and exchange the values of $c$ and $b$ to get a solution with $a > d$. So it is sufficient to assume $a > d$. This makes the left side positive, so the right side must be positive as well. Thus $c > b$.
Now at least one of $c-b$ and $c+b$ must be even. But they differ by $2b$, so both are even. Let $r = \frac{c+b}2, s= \frac{c-b}2$. Then $$(a-d)(a+d) = 2rs$$ So one of $a-d$ and $a+d$ must be even, and thus both. Let $p = \frac{a+d}2, q = \frac{a-d}2$. Now we have $$2pq = rs$$
Solutions to this equation are obvious. Pick arbitrary $p, q$ with $0 < q < p$. Then pick $r > s$ to be any factorization of $2pq$ (including $r = 2pq, s= 1$).
Then set $$a = p+ q\\b=r-s\\c = r+s\\d=p-q$$ to get a solution to $2a^2 + b^2 - c^2 = 2d^2$. Every non-trivial solution to the equation with $a > d$ can be obtained in this way, and every solution with $a < d$ can be obtained by exchanging the values of $(a,c)$ with those of $(d, b)$. The trivial solutions can be obtained by allowing $p=q, r=s$ or $q = s = 0$ as well.