# 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.

You're solving for $k$.

Note by A Former Brilliant Member
1 year, 1 month ago

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@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 $2\in Z^+$ $\Rightarrow k\in Z^+;k>2$

- 1 year, 1 month ago

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

- 1 year, 1 month ago

What about $k \leq 2$? @Zakir Husain

- 1 year, 1 month ago

What I usually do is code these equations. In my output, I got the fact that if you put $x=1,y=1,k=5,$ there are infinite values for $z$. Also, I haven't studied these in school, so I am not at all comfortable in solving using algebra. But there are infinite solutions.

- 1 year, 1 month ago

You're solving for $k$, actually, @Vinayak Srivastava

- 1 year, 1 month ago

But it's not stated. Then we need solutions of the form $(x,y,z,k)$.

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- 1 year, 1 month ago

Sorry, but Algebra is my mortal enemy, I'm very bad at it.

- 1 year, 1 month ago

It is not an algebra but a number theory problem. It is from a branch of number theory called Algebraic number theory

- 1 year, 1 month ago

'Algebraic' Number theory, how's it different?

- 1 year, 1 month ago

There are two broad branches of number theory- Analytic number theory and Algebraic number theory. The difference is that Algebraic number theory uses algebra as a way to get answers to number theory problems like this one. In algebra you will not discriminate between integer, rational, real or complex solutions to the equation but for number theory it matters.

- 1 year, 1 month ago

See algebra is used, so I'm out.

- 1 year, 1 month ago