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.

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.
