Dependent progression

Knowing that $xyz = 55$ , we could start by finding the way to represent $55$ as the product of three numbers. But $55 = 5 \cdot 11$ . So, let's rewrite 55 as an improper fraction we can separate into three factors:

$55 = \frac {110}{2}$ , but we can't separate 2 into three factors.

$55 = \frac {220}{4}$ , but we can't separate 4 into three factors (i'm not considering 1 as a possible factor).

$55 = \frac {440}{8}$ , there's a chance for this option to achieve our goal, as follows:

$55 = \frac {440}{8} = \frac {5}{2} \cdot \frac {8}{2} \cdot \frac {11}{2}$

Of course, the choice of numbers 5,8 and 11 for the decomposition stands for the purpose of constructing the arithmetic progression. Then, the numbers $x$ , $y$ and $z$ form an arithmetic progression of ratio $\frac {3}{2}$ . Knowing that, it's easy to determine the values of $a = \frac {2}{2} = 1$ and $b = \frac {14}{2} = 7$

So, we have a preliminar result: $a^2 + b^2 = 50$ .

Let's see if these values work for the second construction...

Following a similar thinking for the statement $xyz = \frac {343}{55}$ , the idea is to rewrite it as a product of three factors. It's easy to detect that $343 = 7^3$ , so no problem in writing the numerator as $7 \cdot 7 \cdot 7$ . For the denominator, we can use the decomposition obtained above. Thus, the product could be written as:

$\frac {343}{55} = 7 \cdot 7 \cdot 7 \cdot \frac {2}{5} \cdot \frac {2}{8} \cdot \frac {2}{11}$

$\frac {343}{55} = \frac {2 \cdot 7}{5} \cdot \frac {2 \cdot 7}{8} \cdot \frac {2 \cdot 7}{11} = \frac {14}{5} \cdot \frac {14}{8} \cdot \frac {14}{11}$

And these three factors would represent the harmonic progression we were looking for! Now, ordering the terms first, it's easy to determine the values of $a = \frac {14}{14} = 1$ and $b = \frac {14}{2} = 7$ .

And we then confirmed the result: $a^2 + b^2 = 50$