Diophantine Equations (Problem $9$)

$x^y + y^z + z^{x - 1} = k - 2$

Find solutions where $x, y, z, k$ are all positive or negative integers.

Give your answer as the number of solutions.

You're solving for $k$ .

#Algebra

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Comments

@Yajat Shamji- For positive integers $x,y,z,k$ there are infinite solutions:

Take random positive integers $x,y,z$ then $x^y\in Z^+; y^z\in Z^+; z^{x-1}\in Z^+\Rightarrow (x^y+y^z+z^{x-1})\in Z^+$ $\Rightarrow (k-2)\in Z^+$ As $2\in Z^+$ $\Rightarrow k\in Z^+;k>2$

Zakir Husain - 11 months, 3 weeks ago

So you're saying I have made a Diophantine equation that has infinite integer solutions? @Zakir Husain