HMMT February

If any of $H, M, T$ are zero, the product is 0. We can do better (examples below), so we may now restrict attention to the case when $H, M, T \neq 0$. When $M \in\{-2,-1,1,2\}$, a little casework gives all the possible $(H, M, T)=(2,1,4),(4,1,2),(-1,-2,1),(1,-2,-1)$. If $M=-2$, i.e. $H-4+T=4 H T$, then $-15=(4 H-1)(4 T-1)$, so $4 H-1 \in\{ \pm 1, \pm 3, \pm 5, \pm 15\}$ (only $-1,+3,-5,+15$ are possible) corresponding to $4 T-1 \in\{\mp 15, \mp 5, \mp 3, \mp 1\}$ (only $+15,-5,+3,-1$ are possible). But $H, T$ are nonzero, we can only have $4 H-1 \in\{+3,-5\}$, yielding $(-1,-2,1)$ and $(1,-2,-1)$. If $M=+2$, i.e. $H+4+T=4 H T$, then $17=(4 H-1)(4 T-1)$, so $4 H-1 \in\{ \pm 1, \pm 17\}$ (only $-1,-17$ are possible) corresponding to $4 T-1 \in\{ \pm 17, \pm 1\}$ (only $-17,-1$ are possible). But $H, T$ are nonzero, so there are no possibilities here. If $M=-1$, i.e. $H-2+T=H T$, then $-1=(H-1)(T-1)$, so we have $H-1 \in\{ \pm 1\}$ and $T-1 \in\{\mp 1\}$, neither of which is possible (as $H, T \neq 0)$. If $M=+1$, i.e. $H+2+T=H T$, then $3=(H-1)(T-1)$, so we have $H-1 \in\{ \pm 1, \pm 3\}$. Since $H, T \neq 0, H-1 \in\{+1,+3\}$, yielding $(2,1,4)$ and $(4,1,2)$. Now suppose there is such a triple $(H, M, T)$ for $|M| \geq 3$. The equation in the problem gives $\left(M^{2} H-\right.$ 1) $\left(M^{2} T-1\right)=2 M^{3}+1$. Note that since $H, T \neq 0,\left|2 M^{3}+1\right|=\left|M^{2} H-1\right| \cdot\left|M^{2} T-1\right| \geq \min \left(M^{2}-\right.$ $\left.1, M^{2}+1\right)^{2}=M^{4}-2 M^{2}+1>2|M|^{3}+1$ gives a contradiction.