Only figured it out for symmetric triplets...

Assume $$ (a, b, c) = (x - d, x, x + d) $$ for some $$ x $$ and $$ d $$. The sum and product of the triplet are:

- Sum:
$$
(x - d) + x + (x + d) = 3x.
$$

- Product:
$$
(x - d) \cdot x \cdot (x + d) = x \cdot (x^2 - d^2).
$$

We set the sum equal to the product:

$$
3x = x(x^2 - d^2).
$$

Assuming $$ x \neq 0 $$, divide both sides by $$ x $$:

$$
3 = x^2 - d^2.
$$

This leads to the equation:

$$
x^2 = d^2 + 3.
$$

Thus, $$ x = \sqrt{d^2 + 3} $$. For $$ x $$ to be a natural number, $$ d^2 + 3 $$ must be a perfect square.

- For $$ d = 1 $$:
  $$
  x^2 = 1^2 + 3 = 4 \quad \Rightarrow \quad x = 2.
  $$
  The triplet is (x - d, x, x + d) = (1, 2, 3).
  $$
No other values of $$ d $$ yield a perfect square for $$ x^2 $$, because $$ d^2 + 3 $$ is not a perfect square for $$ d > 1 $$.


south_korea_ln

Haven't worked it out fully, but I think it would go something like this:

Suppose $$a+b+c = abc$$.

Then for any $$x,y,z > 0$$:

$$ (a+x)(b+y)(c+z) > (a+x) + (b+y) + (c+z)$$

We'd only have to check $$x,y,z>0$$ because you asked for natural number solutions, and we know 1,2,3 is the smallest such solution.



SimpleStacker

science

[Daily puzzle] log of a sum and sum of logs

That's some lucky trial and error.

Bonus question: can one prove this would be the only solution for $$a,b,c \in \mathbb{N}$$?