Positive Integer Triples $(a,b,c)$ Satisfying $a^3+b^3+c^3=(abc)^2$

[anemone]

Find all triples $(a,\,b,\,c)$ of positive integers such that $a^3+b^3+c^3=(abc)^2$.

 

[jvicens]

After some investigation, I have found that there are only two possible solutions for this equation: (1,1,1) and (2,2,2). Let's take a closer look at why these are the only solutions.

First, we can rearrange the equation to get $a^3+b^3+c^3= (abc)^2 \Rightarrow a^3+b^3+c^3 - (abc)^2 = 0$. This can be factored to $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$.

Since we are looking for positive integers, we can eliminate the first factor of 0. Now we are left with $a^2+b^2+c^2-ab-bc-ca=0$.

We can use the well-known identity $a^2+b^2+c^2 \geq ab+bc+ca$ for all positive integers $a,b,c$ to see that the only way for the left side of the equation to equal 0 is if $a=b=c$. This gives us the solution (1,1,1).

For the second solution, we can use the fact that $a^2+b^2+c^2-ab-bc-ca=0$ is symmetric in its variables. This means that we can swap the values of $a,b,c$ and still get the same result. Therefore, if (a,b,c) is a solution, so are (b,c,a) and (c,a,b). This allows us to generate more solutions by starting with one value and swapping it with the other two. For example, if we start with (1,1,2), we can swap the 2 with the 1's to get (1,2,1) and (2,1,1). All three of these triples will satisfy the equation.

However, upon further inspection, we can see that (1,1,2) and (1,2,1) are not valid solutions since they do not result in positive integers when plugged into the original equation. This leaves us with only one additional solution: (2,2,2).

In conclusion, the only solutions to the equation $a^3+b^3+c^3=(abc)^2$ are (1,1,1) and (2,2,2).