Subfield: Number Theory


Question: Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$.

Solution:
Note that the equation is symmetric. We will assume without loss of generality that $a \geqslant b \geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.


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 $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right) \geqslant a^{2}$, and so

$$
18 b^{3} \geqslant 9\left(b^{3}+c^{3}\right) \geqslant 9 a^{2} \geqslant b^{4} c^{4} \geqslant b^{3} c^{5},
$$

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

$$
4\left(1+b^{3}\right)=4 a^{2}\left(b^{2}-a\right) \geqslant 4 a^{2}>b^{4},
$$

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:

$$
\begin{aligned}
& b=2: \quad a^{3}-4 a^{2}+9=0 \\
& b=3: \quad a^{3}-9 a^{2}+28=0 \\
& b=4: \quad a^{3}-16 a^{2}+65=0
\end{aligned}
$$

The only case with an integer solution for $a$ with $b \leqslant a$ is $b=2$, leading to $(a, b, c)=(3,2,1)$.
Note that the equation is symmetric. We will assume without loss of generality that $a \geqslant b \geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.

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 $(a, b, c)=(3,2,1)$.
Note that the equation is symmetric. We will assume without loss of generality that $a \geqslant b \geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.

One approach to finish the problem after establishing that $c \leqslant 1$ is to set $k=b^{2} c^{2}-a$, which is clearly an integer and must be positive as it is equal to $\left(b^{3}+c^{3}\right) / a^{2}$. Then we divide into cases based on whether $k=1$ or $k \geqslant 2$; in the first case, we have $b^{3}+1=a^{2}=\left(b^{2}-1\right)^{2}$ whose only positive root is $b=2$, and in the second case we have $b^{2} \leqslant 3 a$, and so

$$
b^{4} \leqslant(3 a)^{2} \leqslant \frac{9}{2}\left(k a^{2}\right)=\frac{9}{2}\left(b^{3}+1\right),
$$

which implies that $b \leqslant 4$.

Set $k=\left(b^{3}+c^{3}\right) / a^{2} \leqslant 2 a$, 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

$$
a+k \geqslant\left(k a^{2}-1\right)^{2 / 3}
$$

As proved before, $k$ is a positive integer; for each value of $k \geqslant 1$, this gives us a polynomial inequality satisfied by $a$ :

$$
k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\left(k^{3}-1\right) \leqslant 0
$$

We now prove that $a \leqslant 3$. Indeed,

$$
0 \geqslant \frac{k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\left(k^{3}-1\right)}{k^{2}} \geqslant a^{4}-a^{3}-5 a^{2}-3 a-k \geqslant a^{4}-a^{3}-5 a^{2}-5 a,
$$

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 $(a, b, c)=(3,2,1)$.
Note that the equation is symmetric. We will assume without loss of generality that $a \geqslant b \geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.

Again, observe that $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right)$, so $b \leqslant a \leqslant b^{2} c^{2}-1$.

We consider the function $f(x)=x^{2}\left(b^{2} c^{2}-x\right)$. It can be seen that that on the interval $\left[0, b^{2} c^{2}-1\right]$ 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

$$
b^{3}+c^{3}=f(a) \geqslant \min \left(f(b), f\left(b^{2} c^{2}-1\right)\right)
$$

First, suppose that $b^{3}+c^{3} \geqslant f\left(b^{2} c^{2}-1\right)$. This may be written $b^{3}+c^{3} \geqslant\left(b^{2} c^{2}-1\right)^{2}$, and so

$$
2 b^{3} \geqslant b^{3}+c^{3} \geqslant\left(b^{2} c^{2}-1\right)^{2}>b^{4} c^{4}-2 b^{2} c^{2} \geqslant b^{4} c^{4}-2 b^{3} c^{4}
$$

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

$$
2 b^{3} \geqslant b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right) \geqslant b^{2}\left(b^{2} c^{2}-b\right) .
$$

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.

Final answer: $(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$