HMMT February, 2015

Solution:
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...

Solution 3. The left hand sides clue us in to the fact that this problem is secretly about quaternions. Indeed, we see that letting $z=a+b i+c j+d k$ gives
$$(z-i+j) z=-2-32 i+28 j+31 k$$
Taking norms gives $N(z-i+j) N(z)=2^{2}+32^{2}+28^{2}+31^{2}=2773=47 \cdot 59$. By the triangle inequality, $N(z), N(z-i+j)$ aren't too far apart, so they must be 47,59 (in some order).
Thus $z, z-i+j$ are Hurwitz primes. We rely on the following foundational lemma in quaternion number theory:

Lemma. Let $p \in \mathbb{Z}$ be an integer prime, and $A$ a Hurwitz quaternion. If $p \mid N(A)$, then the $\mathbb{H} A+\mathbb{H} p$ (a left ideal, hence principal) has all element norms divisible by $p$, hence is nontrivial. (So it's either $\mathbb{H} p$ or of the form $\mathbb{H} P$ for some Hurwitz prime $P$.)

In our case, it will suffice to apply the lemma for $A=-2-32 i+28 j+31 k$ at primes $p=47$ and $q=59$ to get factorizations (unique up to suitable left/right unit multiplication) $A=Q P$ and $A=P' Q'$ (respectively), with $P, P'$ Hurwitz primes of norm $p$, and $Q, Q'$ Hurwitz primes of norm $q$. Indeed, these factorizations come from $\mathbb{H} A+\mathbb{H} p=\mathbb{H} P$ and $\mathbb{H} A+\mathbb{H} q=\mathbb{H} Q'$.

We compute by the Euclidean algorithm:
$$\begin{aligned} \mathbb{H} A+\mathbb{H}(47) & =\mathbb{H}(-2-32 i+28 j+31 k)+\mathbb{H}(47) \\ & =\mathbb{H}(-2+15 i-19 j-16 k)+\mathbb{H}(47) \\ & =[\mathbb{H}(47 \cdot 18)+\mathbb{H}(47)(-2-15 i+19 j+16 k)] \frac{-2+15 i-19 j-16 k}{47 \cdot 18} \\ & =[\mathbb{H} 18+\mathbb{H}(-2+3 i+j-2 k)] \frac{-2+15 i-19 j-16 k}{18} \\ & =\mathbb{H}(-2+3 i+j-2 k) \frac{-2+15 i-19 j-16 k}{18} \\ & =\mathbb{H} \frac{-54-90 i+54 j-36 k}{18} \\ & =\mathbb{H}(-3-5 i+3 j-2 k) . \end{aligned}$$
Thus, there's a unit $\epsilon$ such that $P=\epsilon(-3-5 i+3 j-2 k)$.

Similarly, to get $P'$, we compute
$$\begin{aligned} A \mathbb{H}+47 \mathbb{H} & =(-2-32 i+28 j+31 k) \mathbb{H}+47 \mathbb{H} \\ & =(-2+15 i-19 j-16 k) \mathbb{H}+47 \mathbb{H} \\ & =\frac{-2+15 i-19 j-16 k}{47 \cdot 18}[(47 \cdot 18) \mathbb{H}+47(-2-15 i+19 j+16 k) \mathbb{H}] \\ & =\frac{-2+15 i-19 j-16 k}{18}[18 \mathbb{H}+(-2+3 i+j-2 k) \mathbb{H}] \\ & =\frac{-2+15 i-19 j-16 k}{18}(-2+3 i+j-2 k) \mathbb{H} \\ & =\frac{-54+18 i+18 j+108 k}{18} \mathbb{H} \\ & =(-3+i+j+6 k) \mathbb{H}, \end{aligned}$$
so there's a unit $\epsilon'$ with $P'=(-3+i+j+6 k) \epsilon'$.

Finally, we have either $z=\epsilon(-3-5 i+3 j-2 k)$ for some $\epsilon$, or $z-i+j=(-3+i+j+6 k) \epsilon'$ for some $\epsilon'$. Checking the $24+24$ cases (many of which don't have integer coefficients, and can be ruled out immediately) gives $z=i P=5-3 i+2 j+3 k$ as the only possibility.