Equation : x^{n-1} + y^n = z^{n+1}

Let   $n$   be a positive integer.

Prove that there exist distinct positive integers   $x, \; y, \; z$   such that

$x^{n-1} \; + \; y^n \; = \; z^{n+1}$

One solution is :

$x \; = \; 2^{n^2} \: 3^{n+1}$
$y \; = \; 2^{n^2 - n} \: 3^n$
$z \; = \; 2^{n^2 - 2n + 2} \: 3^{n-1}$

math grad - Interest: Number theory
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2 Responses to Equation : x^{n-1} + y^n = z^{n+1}

1. paul says:

Here are a few solutions.
When n=1
{x->1,y->3,z->2}
{x->1,y->8,z->3}
{x->1,y->15,z->4}
{x->1,y->24,z->5}

When n=2
{x->2,y->5,z->3}
{x->4,y->2,z->2}
{x->4,y->11,z->5}
{x->7,y->1,z->2}

When n=3
{x->27,y->18,z->9}
{x->28,y->8,z->6}

Paul.