, 2019

Solution 1. We will start by proving that $c=1$. Note that
$$3 a^{3} \geqslant a^{3} + b^{3} + c^{3} > a^{3}$$
So $3 a^{3} \geqslant (a b c)^{2} > a^{3}$ and hence $3 a \geqslant b^{2} c^{2} > a$. Now $b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a) \geqslant a^{2}$, and so
$$18 b^{3} \geqslant 9 (b^{3} + c^{3}) \geqslant 9 a^{2} \geqslant b^{4} c^{4} \geqslant b^{3} c^{5}$$
so $18 \geqslant c^{5}$ which yields $c=1$.

Now, note that we must have $a > b$, as otherwise we would have $2 b^{3} + 1 = b^{4}$ which has no positive integer solutions. So
$$a^{3} - b^{3} \geqslant (b + 1)^{3} - b^{3} > 1$$
and
$$2 a^{3} > 1 + a^{3} + b^{3} > a^{3},$$
which implies $2 a^{3} > a^{2} b^{2} > a^{3}$ and so $2 a > b^{2} > a$. Therefore
$$4 (1 + b^{3}) = 4 a^{2} (b^{2} - a) \geqslant 4 a^{2} > b^{4}$$
so $4 > b^{3} (b - 4)$; that is, $b \leqslant 4$.

Now, for each possible value of $b$ with $2 \leqslant b \leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows:
$$\begin{array}{ll} b=2: & a^{3} - 4 a^{2} + 9 = 0 \\ b=3: & a^{3} - 9 a^{2} + 28 = 0 \\ b=4: & a^{3} - 16 a^{2} + 65 = 0 . \end{array}$$
The only case with an integer solution for $a$ with $b \leqslant a$ is $b=2$, leading to $(a, b, c) = (3, 2, 1)$.

Solution 2. Again, we will start by proving that $c=1$. Suppose otherwise that $c \geqslant 2$. We have $a^{3} + b^{3} + c^{3} \leqslant 3 a^{3}$, so $b^{2} c^{2} \leqslant 3 a$. Since $c \geqslant 2$, this tells us that $b \leqslant \sqrt{3 a} / 4$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \leqslant 2 b^{3} \leqslant 2 (3 a / 4)^{3 / 2}$. In other words, $a \leqslant \frac{27}{16} < 2$, which contradicts the assertion that $a \geqslant c \geqslant 2$. So there are no solutions in this case, and so we must have $c=1$.

Now, the original equation becomes $a^{3} + b^{3} + 1 = a^{2} b^{2}$. Observe that $a \geqslant 2$, since otherwise $a = b = 1$ as $a \geqslant b$.
The right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3} + 1 \geqslant a^{2}$. Since $a \geqslant b$, we also have
$$b^{2} = a + \frac{b^{3} + 1}{a^{2}} \leqslant 2 a + \frac{1}{a^{2}}$$
and so $b^{2} \leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2} + 1 \geqslant b^{3} + 1 \geqslant a^{2}$, from which we deduce $a \leqslant 8$.

Now, for each possible value of $a$ with $2 \leqslant a \leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows:
$$\begin{array}{ll} a=2: & b^{3} - 4 b^{2} + 9 = 0 \\ a=3: & b^{3} - 9 b^{2} + 28 = 0 \\ a=4: & b^{3} - 16 b^{2} + 65 = 0 \\ a=5: & b^{3} - 25 b^{2} + 126 = 0 \\ a=6: & b^{3} - 36 b^{2} + 217 = 0 \\ a=7: & b^{3} - 49 b^{2} + 344 = 0 \\ a=8: & b^{3} - 64 b^{2} + 513 = 0 . \end{array}$$
The only case with an integer solution for $b$ with $a \geqslant b$ is $a=3$, leading to $(a, b, c) = (3, 2, 1)$.

Solution 3. Set $k = (b^{3} + c^{3}) / a^{2} \leqslant 2 a$, and rewrite the original equation as $a + k = (b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \geqslant b^{3} + c^{3} - 1 = k a^{2} - 1$, so
$$a + k \geqslant (k a^{2} - 1)^{2 / 3}$$
As in Comment 1.2, $k$ is a positive integer; for each value of $k \geqslant 1$, this gives us a polynomial inequality satisfied by $a$:
$$k^{2} a^{4} - a^{3} - 5 k a^{2} - 3 k^{2} a - (k^{3} - 1) \leqslant 0 .$$
We now prove that $a \leqslant 3$. Indeed,
$$0 \geqslant \frac{k^{2} a^{4} - a^{3} - 5 k a^{2} - 3 k^{2} a - (k^{3} - 1)}{k^{2}} \geqslant a^{4} - a^{3} - 5 a^{2} - 3 a - k \geqslant a^{4} - a^{3} - 5 a^{2} - 5 a,$$
which fails when $a \geqslant 4$.
This leaves ten triples with $3 \geqslant a \geqslant b \geqslant c \geqslant 1$, which may be checked manually to give $(a, b, c) = (3, 2, 1)$.

Solution 4. Again, observe that $b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a)$, so $b \leqslant a \leqslant b^{2} c^{2} - 1$.
We consider the function $f(x) = x^{2} (b^{2} c^{2} - x)$. It can be seen that that on the interval $[0, b^{2} c^{2} - 1]$ the function $f$ is increasing if $x < \frac{2}{3} b^{2} c^{2}$ and decreasing if $x > \frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that
$$b^{3} + c^{3} = f(a) \geqslant \min \left(f(b), f(b^{2} c^{2} - 1)\right)$$
First, suppose that $b^{3} + c^{3} \geqslant f(b^{2} c^{2} - 1)$. This may be written $b^{3} + c^{3} \geqslant (b^{2} c^{2} - 1)^{2}$, and so
$$2 b^{3} \geqslant b^{3} + c^{3} \geqslant (b^{2} c^{2} - 1)^{2} > b^{4} c^{4} - 2 b^{2} c^{2} \geqslant b^{4} c^{4} - 2 b^{3} c^{4}$$
Thus, $(b - 2) c^{4} < 2$, and the only solutions to this inequality have $(b, c) = (2, 2)$ or $b \leqslant 3$ and $c = 1$. It is easy to verify that the only case giving a solution for $a \geqslant b$ is $(a, b, c) = (3, 2, 1)$.

Otherwise, suppose that $b^{3} + c^{3} = f(a) \geqslant f(b)$. Then, we have
$$2 b^{3} \geqslant b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a) \geqslant b^{2} (b^{2} c^{2} - b)$$
Consequently $b c^{2} \leqslant 3$, with strict inequality in the case that $b \neq c$. Hence $c = 1$ and $b \leqslant 2$. Both of these cases have been considered already, so we are done.