HMMT February

Solution:
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 $-15 = (4H - 1)(4T - 1)$, so $4H - 1 \in \{ \pm 1, \pm 3, \pm 5, \pm 15 \}$ (only $-1, +3, -5, +15$ are possible) corresponding to $4T - 1 \in \{\mp 15, \mp 5, \mp 3, \mp 1\}$ (only $+15, -5, +3, -1$ are possible). But $H, T$ are nonzero, we can only have $4H - 1 \in \{+3, -5\}$, yielding $(-1, -2, 1)$ and $(1, -2, -1)$.

- If $M = +2$, i.e. $H + 4 + T = 4 H T$, then $17 = (4H - 1)(4T - 1)$, so $4H - 1 \in \{ \pm 1, \pm 17 \}$ (only $-1, -17$ are possible) corresponding to $4T - 1 \in \{ \pm 17, \pm 1 \}$ (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 $(M^{2} H - 1)(M^{2} T - 1) = 2 M^{3} + 1$. Note that since $H, T \neq 0$, $|2 M^{3} + 1| = |M^{2} H - 1| \cdot |M^{2} T - 1| \geq \min(M^{2} - 1, M^{2} + 1)^{2} = M^{4} - 2 M^{2} + 1 > 2|M|^{3} + 1$ gives a contradiction.