, 2023

We first rule out the possibility that $a = b$. Note that if $a = b$, then for every prime $p \mid a$, since $p \mid ab$ and $ab$ is a perfect square, we have $p^2 \mid ab$, which implies $p \mid a+b-3\sqrt[3]{abc}$. This means $p \nmid c$, and hence $\gcd(a, c) = 1$. Moreover, since $\sqrt[3]{abc}$ is an integer, there must exist $m, n \in \mathbb{N}$ such that $(a, b, c) = (m^3, m^3, n^3)$, so that
$$a + b + c - 3\sqrt[3]{abc} = 2m^3 + n^3 - 3m^2n = (m - n)^2(2m + n) = 1,$$
which clearly has no solution, so $a \neq b$.

Next, without loss of generality assume $a > b$. By the AM-GM inequality, we have
$$\begin{aligned} 1 &= a + b + c - 3\sqrt[3]{abc} \\ &= (\sqrt{a} - \sqrt{b})^2 + \left(\sqrt{ab} + \sqrt{ab} + c - 3\sqrt[3]{\sqrt{ab}\sqrt{abc}}\right) \\ &\ge (\sqrt{a} - \sqrt{b})^2 > 0, \end{aligned}$$
hence $\sqrt{a} - \sqrt{b} = 1$. Moreover, since $ab$ is a perfect square, $a$ and $b$ are both perfect squares$^1$, so there exists $n \in \mathbb{N}$ such that $a = (n+1)^2$ and $b = n^2$. Also, since equality must hold in the AM-GM inequality above, we have $c = \sqrt{ab} = n(n+1)$. Substituting the above solution back confirms that it holds. This completes the proof.

$^1$For instance, via the identity $\sqrt{a} = (a - b + 1)/2$