Find all positive integer $a,b,c$ that $a^3+b^3+c^3$ can be divided by $a^2b,b^2c,c^2a$

Question

Find all triplets of positive integers $(a,b,c)$ for which $$a^3+b^3+c^3$$ is divisible by $a^2b$, $b^2c$ and $c^2a$.

I just found that $a=b=c$ satisfies the problem. Are there any other possible answers?

How to find all possible answers?

Answer 1

Let $a$, $b$ and $c$ be positive integers for which $a^3+b^3+c^3$ is divisible by $a^2b$, $b^2c$ and $c^2a$. Then $\gcd(a,b)^3$ divides $a^2b$, $a^3$ and $b^3$, and hence also $$(a^3+b^3+c^3)-a^3-b^3=c^3,$$ which shows that $\gcd(a,b)$ divides $c$, so $\gcd(a,b,c)=\gcd(a,b)$. By symmetry $$\gcd(a,b,c)=\gcd(a,b)=\gcd(a,c)=\gcd(b,c),$$ and dividing $a$, $b$ and $c$ by this common factor yields another solution, so we may assume without loss of generality that $\gcd(a,b,c)=1$. Then it follows that $a^3+b^3+c^3$ is divisible by $$\gcd(a^2b,b^2c,c^2a)=a^2b^2c^2.$$ Also without loss of generality $a\leq b\leq c$, and so $$a^2b^2c^2\leq a^3+b^3+c^3\leq3c^3,$$ which shows that $a^2b^2\leq3c$. In particular, because $c^2$ divides $a^3+b^3+c^3$ we also have $$c^2\leq a^3+b^3\leq2b^3,$$ and putting this together with $b^2\leq a^2b^2\leq3c$ we find that $$c^4\leq4b^6\leq108c^3,$$ and hence that $c\leq108$. From here it is easy to verify that $(1,1,1)$ and $(1,2,3)$ are the only valid triplets (up to permutation and scaling) though without the help of a computer this may take some time.

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To verify this manually, note that from $a^2b^2\leq3c$ and $a\leq b$ and $c\leq108$ we get $$b\leq\frac{\sqrt{3c}}{a}\qquad\text{ and }\qquad a\leq\sqrt[4]{3c}<5.$$

For $a=4$ we have $b\leq\tfrac{\sqrt{3c}}{a}<5$ and so from $a\leq b$ we get $b=4$. This contradicts $\gcd(a,b)=1$.

For $a=3$ we have $b\leq\frac{\sqrt{3c}}{a}\leq6$. From $\gcd(a,b)=1$ it follows that $b\in\{4,5\}$ and so either $$a^3+b^3=7\times13\qquad\text{ or }\qquad a^3+b^3=2^3\times19.$$ Because $c^2$ divides $a^3+b^3$ and $c\geq b$ we again reach a contradiction.

For $a=2$ we have $b\leq\frac{\sqrt{3c}}{a}\leq9$ . Then $b\in\{3,5,7,9\}$ and so the the factorizations $$2^3+3^3=3\times7\qquad 2^3+5^3=7\times19\qquad 2^3+7^3=3^3\times13\qquad 2^3+9^3=11\times67,$$ again yield a contradiction with the facts that $c\geq b$ and $c^2$ divides $a^3+b^3$.

For $a=1$ we have $b\leq\frac{\sqrt{3c}}{a}\leq18$. Clearly $(a,b,c)=1$ is a solution, and for any other solution we have $c\geq2$ and $c>b$. We know that $c^2$ divides $$a^3+b^3=1+b^3=(1+b)(b^2-b+1),$$ and computing the factorizations for all $b\leq18$ shows that this product is divisible by a square greater than $b^2$ only if $b=2$, and then $c=3$.

Answer 2

There are other solutions.

E.g. $(a, b, c) = (k, 2k, 3k), (k, 3k, 2k)$ (which are different up to symmetry).