reply on: [Daily puzzle] log of a sum and sum of logs \ stacker news ~science

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trial and error: a=1 b=2 c=3 I don't remember the log rules anymore :)

Or any combination of 1, 2, and 3.

You get it because:

\log(a) + \log(b) + \log(c) = \log(abc)

Thus, the equation is the same as:

a + b + c = abc

, with

a, b, c > 0

👍

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}

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.

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:

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

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

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