Solution 1. Subtract the second equation from the third to get $a(c-b+1)=30$. Add the second and third to get $2 a(b+c)=-4-2 d$. Substitute into the fourth to get $2 a(2 a d-31)=-4-2 d \Longleftrightarrow a(31-2 a d)=2+d \Longleftrightarrow d=\frac{31 a-2}{2 a^{2}+1}$ which in particular gives $a \not \equiv 1(\bmod 3)$. Then plugging in a factor of 30 for $a$ gives us the system of equations $b+c=2 a d-31$ and $c-b+1=30 / a$ in $b, c$. Here, observe that $b+c$ is odd, so $c-b+1$ is even. Thus $a$ must be odd (and from earlier $a \not \equiv 1(\bmod 3)$ ), so $a \in\{-1, \pm 3,5, \pm 15\}$. Manually checking these, we see that the only possibilities we need to check are $(a, d)=(5,3),(-1,-11),(-3,-5)$, corresponding to $(b, c)=(-3,2),(11,-20),(5,-6)$. Then check the three candidates against first condition $a^{2}-b^{2}-c^{2}-d^{2}=c-b-2$ to find our only solution $(a, b, c, d)=(5,-3,2,3)$. Solution 2. Here's an alternative casework solution. From $2 a d=b+c+31$, we have that $b+c$ is odd. So, $b$ and $c$ has different parity. Thus, $b^{2}+c^{2} \equiv 1(\bmod 4)$. Plugging this into the first equation, we get that $a$ and $d$ also have the same parity. So, $a^{2}-b^{2}-c^{2}-d^{2} \equiv-1(\bmod 4)$. Thus, $c-b-2 \equiv-1(\bmod 4)$. So, $c \equiv b+1(\bmod 4)$. From taking modulo $a$ in the second and third equation, we have $a \mid d+32$ and $a \mid 28-d$. So, $a \mid 60$. Now, if $a$ is even, let $a=2 k$ and $d=2 m$. Plugging this in the second and third equation, we get $2 k c=14-k-m$ and $2 k b=k-m-16$. So, $k(c-b)=15-k$. We can see that $k \neq 0$. Therefore, $c-b=\frac{15-k}{k}=\frac{15}{k}-1$. But $c-b \equiv 1(\bmod 4)$. So, $\frac{15}{k}-1 \equiv 1(\bmod 4)$, or $\frac{15}{k} \equiv 2(\bmod 4)$ which leads to a contradiction. So, $a$ is odd. And we have $a \mid 60$. So, $a \mid 15$. This gives us 8 easy possibilities to check...