A much more formal way to do this

Question

The number of positive integer $a,b,c$ satisfying $$a^{b^c}b^{c^a}c^{a^b}=5abc$$ is __

$$a^{b^{c}-1}b^{c^{a}-1}c^{a^{b}-1}=5$$ Since five is prime number so the possible value$a^{b^{c}-1},b^{c^{a}-1},c^{a^{b}-1}$ become $1,1,5$ or $1,5,1$ or $5,1,1$.

Now,I am only solving for $1,1,5$ $$a^{b^{c}-1}=1\qquad b^{c^{a}-1}=1 \qquad c^{a^{b}-1}=5 $$ Now solving these will be yield me satisfying integers.

I just want to know, is there any other formal way to do so? Above work is just for that I have tried the question.

Answer 1

If $a,b,c\geq2$, then $a^{b^c} \geq a^4 \geq 8a$, $b^{c^a} \geq b^4 \geq 8b$, $c^{a^b} \geq c^4 \geq 8c$, so $$a^{b^c}b^{c^a}c^{a^b} \geq 512abc > 5abc$$

So $a=1$, $b=1$, or $c=1$. You can handle these cases easily.

I'll just do $c=1$ (rest goes similarily), which gives: $a^bb=5ab$, so $a^b=5a$, which means that $a=5$ and $b=2$.

Your work is fine, by the way.