#
Problem
of the Week

Problem
D and Solution

The
Same Power

## Problem

Sometimes two powers that are not written with the same base are
still equal in value. For example, \(9^3 =
27^2\) and \((-5)^4=25^2\).

If \(x\) and \(y\) are integers, find all ordered pairs
\((x,y)\) that satisfy the equation
\[(x-1)^{x+y}=8^2\]

## Solution

Since \(8^2 = 64\), we want to look
at how we can express \(64\) as \(a^b\) where \(a\) and \(b\) are integers. There are six ways to do
so. We can do so as \(64^1\), \(8^2\), \(4^3\), \(2^6\), \((-2)^6\), and \((-8)^2\).

We use these powers and the expression \((x-1)^{x+y}\) to find values for \(x\) and \(y\).

The power \((x-1)^{x+y}\) is
expressed as \(64^1\) when \(x-1 = 64\) and \(x+y = 1\). Then \(x=65\) and \(y=-64\) follows. Thus \((65,-64)\) is one pair.

The power \((x-1)^{x+y}\) is
expressed as \(8^2\) when \(x-1 = 8\) and \(x+y=2\). Then \(x=9\) and \(y=-7\) follows. Thus \((9,-7)\) is one pair.

The power \((x-1)^{x+y}\) is
expressed as \(4^3\) when \(x-1 = 4\) and \(x+y = 3\). Then \(x=5\) and \(y=-2\) follows. Thus \((5,-2)\) is one pair.

The power \((x-1)^{x+y}\) is
expressed as \(2^6\) when \(x-1 = 2\) and \(x+y = 6\). Then \(x=3\) and \(y=3\) follows. Thus \((3,3)\) is one pair.

The power \((x-1)^{x+y}\) is
expressed as \((-2)^6\) when \(x-1 = -2\) and \(x+y = 6\). Then \(x=-1\) and \(y=7\) follows. Thus \((-1,7)\) is one pair.

The power \((x-1)^{x+y}\) is
expressed as \((-8)^2\) when \(x-1 = -8\) and \(x+y = 2\). Then \(x=-7\) and \(y=9\) follows. Thus \((-7,9)\) is one pair.

Therefore, there are six ordered pairs that satisfy the
equation.

They are \((65,-64)\), \((9,-7)\), \((5,-2)\), \((3,3)\), \((-1,7)\), and \((-7,9)\).