Question

Find all the positive integers x, y, z satisfying the condition x^y^z * y^z^x * z^x^y = 7xyz

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, x, y, z are positive integers and

$\rm \: {x}^{ {y}^{z} }. {y}^{ {z}^{x} }. {z}^{ {y}^{x} } = 7xyz$

can be rewritten as

$\rm \: \dfrac{{x}^{ {y}^{z} }. {y}^{ {z}^{x} }. {z}^{ {y}^{x} }}{xyz} = 7$

$\rm \: \dfrac{{x}^{ {y}^{z} }}{x} \times \dfrac{{y}^{ {z}^{x} }}{y} \times \dfrac{{z}^{ {y}^{x} }}{z} = 7$

$\rm \: {x}^{ {y}^{z} - 1} \times {y}^{ {z}^{x} - 1} \times {z}^{ {y}^{x} - 1} = 7$

As 7 is a prime number, so three cases arises.

Case :- 1

$\rm \: {x}^{ {y}^{z} - 1} = 7 \: \: or \: \: {y}^{ {z}^{x} - 1} = 1 \: \: or \: \: {z}^{ {y}^{x} - 1} = 1$

$\rm\implies \:x = 7 \: \: and \: \: {y}^{z} - 1 = 1$

$\rm\implies \:x = 7 \: \: and \: \: {y}^{z} = 2$

$\rm\implies \:x = 7 \: \: and \: \: y = 2 \: \: and \: \: z = 1$

Case :- 2

$\rm \: {x}^{ {y}^{z} - 1} = 1 \: \: or \: \: {y}^{ {z}^{x} - 1} = 7 \: \: or \: \: {z}^{ {y}^{x} - 1} = 1$

$\rm\implies \:y = 7 \: \: and \: \: {z}^{x} - 1 = 1$

$\rm\implies \:y = 7 \: \: and \: \: {z}^{x} = 2$

$\rm\implies \:y = 7 \: \: and \: \: z = 2 \: \: and \: \: x = 1$

Case :- 3

$\rm \: {x}^{ {y}^{z} - 1} = 1 \: \: or \: \: {y}^{ {z}^{x} - 1} = 1 \: \: or \: \: {z}^{ {y}^{x} - 1} = 7$

$\rm\implies \:z = 7 \: \: and \: \: {x}^{y} - 1 = 1$

$\rm\implies \:z = 7 \: \: and \: \: {x}^{y} = 1 + 1$

$\rm\implies \:z = 7 \: \: and \: \: {x}^{y} = 2$

$\rm\implies \:z = 7 \: \: and \: \: x = 2 \: \: and \: \: y = 1$

Hence, all possible positive integers satisfied the given expression

$\rm \: {x}^{ {y}^{z} }. {y}^{ {z}^{x} }. {z}^{ {y}^{x} } = 7xyz \: \: are$

$\begin{array}{|c|c|c|} \hline \rm x&\rm y&\rm z \\ \hline\rm 7&\rm 2&\rm 1 \\ \hline \rm 1&\rm 7&\rm 2 \\ \hline \rm 2&\rm 1&\rm 7 \\ \hline \end{array}$