Relationships between the elements $(a,b,c,d)$ of a solution to $A^2+B^2+4=C^2+D^2$ I have reduced a certain equation (in positive integers) to the equation
$$A^2 + B^2 + 4 = C^2 + D^2. \quad(\star)$$
Assume the positive integers $(a,b,c,d)$ are any solution to $(\star)$. Are there any algebraic restrictions on [i.e., relationships between] the elements which are well-known or easy to prove? I'm thinking of things like Pythagorean means, or other relative size or congruence restrictions.
Thank you,
Kieren.
 A: A subset of integer solutions to,
$$A^2+B^2+4 = C^2+D^2\tag{1}$$
can be reduced to solving the Pell equation $x^2-ny^2 = 1$ for any non-square $n$ (where $n=5$ will involve the Fibonacci numbers). Consider the more general,
$$x_1^2+x_2^2+x_3^2 = y_1^2+y_2^2+y_3^2\tag{2}$$
The complete solution is,
$$(a+b)^2+(c+d)^2+(e+f)^2 = (a-b)^2+(c-d)^2+(e-f)^2\tag{3}$$
where,
$$ab+cd+ef = 0\tag{4}$$
It is easy to make the last term of $(3)$ vanish. Let $e=f=t$. Thus,
$$(a+b)^2+(c+d)^2+(2t)^2 = (a-b)^2+(c-d)^2\tag{5}$$
However, if you set $a,b,c = x+y,\, -x+y,\, y^2$, then condition $(4)$ becomes,
$$x^2-(d+1)y^2 = t^2\tag{6}$$
If you set $t = 1$, then any Pell equation will give integer solutions to $(1)$.  For a nice example, let $d = 4$ so one has to solve,
$$x^2-5y^2 = 1$$
The solution is $x_n = \tfrac{1}{2}F_{6n}+F_{6n-1}$ (A023039) and $y_n = \tfrac{1}{2}F_{6n}$ (A060645), where $F_n$ are the Fibonacci numbers. After some tweaking so that all terms won't be even, a Fibonacci identity which satisfies $(1)$ is then,
$$(F_{6n+3})^2+( \tfrac{1}{4}F_{6n+3}^2+4)^2+4 = (F_{6n+2}+F_{6n+4})^2+(\tfrac{1}{4}F_{6n+3}^2-4)^2\tag{7}$$
where $n=1$ is,
$$34^2+293^2+4 = 76^2 +285^2$$
and so on.
A: One can pick up some things from congruence considerations. Here is a sample. Let us work modulo $8$.
If $a$ and $b$ are odd, the left side is congruent to $6$ modulo $8$. But the right side cannot be congruent to $6$ modulo $8$. (We used the fact that if $x$ is odd, then $x^2\equiv 1\pmod{8}$.) 
Suppose now that $a$ is even and $b$ is odd. If $a$ is divisible  by $4$, then $a^2+b^2+4$ is congruent to $5$ modulo $8$, which means that one of $c$ or $d$ is odd, and the other is even but not divisibl by $4$.
Let $a$ and $b$ be both even. Then $c$ and $d$ also must be. We end up with an equation of shape $s^4+t^4 +1=u^2+v^2$. 
We could also pick up information by working modulo $3$, and other moduli. 
A: I'm not sure you can say a great deal. If $A^2+B^2=C^2$ (pythagoren triple), then we can have $D=2$.
We have $4=(C+A)(C-A)+(D+B)(D-B)$ so that $C+A$ and $D+B$ are both odd or both even.
In the even case, both products are divisible by $4$ and we have some expression like $4=56-52=14\times 4 - 26\times 2$ which gives $C=9, A=5, B=14, D=12$
In the odd case we get something like $4=55-51=5\times 11-3\times 17$ which gives $C=8, A=3, D=7, B=10$
A: The conditions for a number $n$ to be a sum of two squares are well-understood: $n$ must have no factors congruent to $3\bmod 4$ which occur to odd powers; in other words, if $p\equiv 3\pmod 4, p^{2k+1}|n$, then $p^{2k+2}|n$ (and in particular, if $p|n$ then $p^2|n$). This is a direct consequence of Fermat's Two-Square Theorem and the rule $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$ for expressing the product of two numbers, each a sum of two squares, as a sum of two squares.
Applied to the given problem, this says that both $n=A^2+B^2$ and $n+4$ have no factors of their squarefree parts congruent to $3\bmod 4$; unfortunately, since addition plays notoriously poorly with factorization (for instance, beyond the trivial lack of shared factors, there's essentially nothing known about any correlation between the factorizations of $n$ and $n+1$) then it's unlikely that you'll be able to say anything about $A,B,C,D$ aside from the simplest congruence implications.
A: For the equation:  $X^2+Y^2+4=Z^2+D^2$
You can draw a simple formula.
$X=(2c+b+2)c+b$
$Y=(2c+b+2)c$
$Z=2c^2+bc-2$
$D=(2c+b+4)c+b$
more:
$X=\frac{(a^2-q^2)b}{2}+(b-2)(q-1)$
$Y=ab$
$Z=bq+2-b$
$D=\frac{(a^2-q^2)b}{2}+(b-2)q+2$
more:
$X=\frac{(a^2-q^2)b}{2}-(b+2)(q+1)$
$Y=ab$
$Z=b+bq+2$
$D=\frac{(a^2-q^2)b}{2}-(b+2)q-2$
$c,b,q,a$ - what some integers.
