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

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.

$a^2=a1^2+a2^2-a3^2-a4^2\tag{1}$
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}