Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$. 

Prove that:

For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$

After many hours, I found a general formula satisfying the statement for all $A$ and $B$. $$A^2=(3A+B)^2+(9A+2B)^2-(5A+B)^2-(8A+2B)^2$$ This was derived by noticing the following pattern, to which I used the first equation below as a seed and multiplied through $A^2$. I was then able to find a second paramater $B$ due to the arithmetic progression of the pattern, in order to prove the infinitude of $\{a_n\}_{n=1}^{4}$. $$\begin{align}1^2+5^2+8^2 &= 3^2+9^2 \\ 1^2+6^2+10^2 &= 4^2+11^2 \\ 1^2+7^2+12^2 &= 5^2+13^2 \\ &\vdots\end{align}$$
My question is, is there a way one could prove the statement without the involvement of such curious patterns surrounding square numbers? Apologies if this question is somewhat vague.

Edit:
Fun fact, it appears there is also a general formula for the equation $$A=a_1^2+a_2^2+a_3^2-a_4^2+a_5^2$$ That is, $$A^2=(A+B)^2+(A+3B)^2+(A+8B)^2-(A+5B)^2-(A+7B)^2$$
Edit 2: It appears that the first general equation at which I arrived in this question is actually part of an even more general equation $$(pq +s)^2=\big\{p(3q+r)+3s\big\}^2+\big\{p(9q+r)+3s\big\}^2-\big\{p(5q+r)+s\big\}^2-\big\{2p(4q+r)+4s\big\}^2+4pqs$$ where $(p,q,r,s)=(1,A,B,0)$. Note that an interesting fact implies when $p$, $q$ and $s$ are square numbers.
 A: Note that 
$$a_1^2+a_2^2-a_3^2-a_4^2=\underbrace{(a_1+a_3)(a_1-a_3)}_{=:M}+\underbrace{(a_2+a_4)(a_2-a_4)}_{=:N} $$
where $M$ and $N$ can be any integer that is either odd or a multiple of $4$. In particular, negatives are allowed so that for each number ($A^2$ or otherwise), we find infinitely many such $M,N$.
Concretely, let $C=A^2$ (which in fact need not really be a perfect square).
Pick $R,S$ with $R>S>\max\{2,C\}$ and $R\equiv S\not\equiv C\pmod 2$. Let  $N=RS-C$. Note that $N$ is odd and $>2R$.
Now let 
$$a_1=\frac{R+S}2,\quad a_3=\frac{R-S}2, a_2=\frac{N-1}2, a_4=\frac{N+1}2. $$
Then
$$\begin{align} a_1^2+a_2^2-a_3^2-a_4^2&=(a_1+a_3)(a_1-a_3)-(a_2+a_4)(a_4-a_2)\\&=RS-N=C,\end{align}$$
as desired. 
Also,
$$a_3<a_1<R\le a_2<a_4,$$
i.e., the numbers are distinct.
A: $a^2=a1^2+a2^2-a3^2-a4^2\tag{1}$
Let assume $p^2+q^2-r^2-s^2 = 1\tag{2}$
Equation $(2)$ has many parametric solutions.
We use one of the solutions,  $(p,q,r,s)=(2n+1, n-1, n+1, 2n).$
( By consider two identies, $(n+1)^2 - (n-1)^2 = 4n, (2n+1)^2 - (2n)^2 = 4n+1$ )
$n$ is arbitrary.  
Substitute  $a1=pt+c, a2=qt+d, a3=rt+c, a4=st+d, a=t$  to equation $(1)$, then we get  $$c=s-q$$ $$d=p-r.$$
Thus, we get a parametric solution below.
\begin{eqnarray}
  &a& = t \\ 
  &a1& = (2n+1)t+n+1 \\ 
  &a2& = (n-1)t+n  \\ 
  &a3& = (n+1)t+n+1 \\  
  &a4& = 2nt+n \\   
  \end{eqnarray}
$t$ is arbitrary.  
Example:
\begin{eqnarray}
&(t)^2& = (3t+2)^2 + (1)^2 - (2t+2)^2 - (2t+1)^2  \\ 
&(t)^2& = (5t+3)^2 + (t+2)^2 - (3t+3)^2 - (4t+2)^2  \\ 
&(t)^2& = (7t+4)^2 + (2t+3)^2 - (4t+4)^2 - (6t+3)^2  \\ 
&(t)^2& = (9t+5)^2 + (3t+4)^2 - (5t+5)^2 - (8t+4)^2  \\ 
&(t)^2& = (11t+6)^2 + (4t+5)^2 - (6t+6)^2 - (10t+5)^2  \\ 
\end{eqnarray}
A: Given any non-negative integer $A$, let $a_1=A+2n+1$ for any of $n=1,2,3,\cdots$.
For convenience let $b=a_1^2-A^2 = (A+2n+1)^2-A^2= (4n+2)A+4n^2+4n+1$.  Thus $b$ is odd and $\geq9$.
Now let $a_2=(b-1)/2$ and $a_3=(b+1)/2$, implying that $a_2^2-a_3^2=-b$, and let $a_4=0$. Then:
$$a_1^2+a_2^2-a_3^2-a_4^2=a_1^2+(-b)-0=b+A^2-b=A^2$$
Also $a_3=a_2+1>a_2$, $a_1=A+2n+1>0=a_4$, and: $$a_2=(b-1)/2=(2n+1)A+2n^2+2n>A+2n+1=a_1$$ so the four terms are all unequal.
For negative $A$, let $a_1=|A|+2n+1$ and proceed in the same way.
