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### Unsolvable Equations

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Date: 11/10/2001 at 03:48:38
From: Shawn S.
Subject: Equations in the form x^n+y^n=z

I've been stuck on this problem.  If I have an equation in the form of
x^n + y^n = z, how do I solve for n?

For example, in 3^n + 4^n = 25 I know that n = 2, but how can I find
that without guessing and checking?  I can isolate one of the n's at a
time, but can't figure out how to get both of 'em.  I know about
logarithms and all that.

Here's what I've done:

I got 1 of the n's alone:
n = log(base 3)(25-4^n)

but the other one is still there, so I finally thought of doing this:
n = log(b3)(25-4^log(b3)(25-4^log(b3)(25-4^log(b3)(25-4^...))))

I was replacing the 4^n with 4^log(b3)(25-4^n) and so on, since n =
log(base 3)(25-4^n). But now I have an infinitely long equation.  I
tried calculating it for each step starting with n = 4^log(b3)(25-4),
then n = 4^log(b3)(25-4^log(b3)(25-4))and so on. But to my suprise I
got something like this:  2.3, 3.6, 4.8... It was getting farther away
from 2.  Can you help?
```

```
Date: 11/10/2001 at 07:00:35
From: Doctor Mitteldorf
Subject: Re: Equations in the form x^n+y^n=z

Dear Shawn,

Congratulations to you! You are discovering math on your own, which is
by far the best way to learn. Nobody can fool you about what's true
and what's false, because you're going to try it yourself.  Keep going
with your endeavors, and here's what I think you'll be finding:

1) Solving for n.  When you take an algebra course, teachers will
concentrate on how to solve this kind of equation and how to solve
that kind of equation. But many of them won't bother to tell you that
most equations can't be solved at all. Even an equation as simple as

x + log(x) = 6

cannot be solved for x. You discovered that another form with this
property:

x^n + y^n = z

cannot be solved for n.

But now I'm going to get philosophical on you. Let's take the equation

x^2 = 3

This one does have a solution, you say: x = sqrt(3). But once upon a
time, people didn't know about square roots, so this equation did not
have a solution. The way they made it a "solvable" equation was to
define a function called square root that was the solution to the
equation. Then they calculated square roots (by guess-and-correct) so
many times that they started to feel as though square root was an old
friend. There were tables of square roots, and different algorithms
for calculating the square root by hand, or calculating it on a
computer.

Maybe the difference between a "solvable" equation and an "unsolvable"
equation is not so much about the equation as about people and
history. Some equations have been studied and functions invented and
named, while others have not. Besides the square root, can you think
of other examples that illustrate this point?

2) Once you realize that you can't just "solve for n," you will need
to create an algorithm for guess-and-check that gets you ever closer
to the solution. You have a good idea for creating an algorithm: Just
solve for n on the left side of the equation. It's okay if n appears
on the right side. Guess the n for the right side, use it to calculate
the n on the left, then start all over again and use that calculated n

Starting with the equation 3^n + 4^n = 25, you could subtract 4^n from
both sides, then take the log (base 10 or base e or whatever, it
doesn't matter)

n log(3) = log(25 - 4^n)

Solving for n, you find

n = log(25-4^n)/log(3)

If you guess the first value of n, and put this on the right, then
solve for the n on the left and use that value for your next guess,
you find that the answers don't get closer and closer to 2 - they get
farther and farther away.

But you might have started, instead, with

n log(4) = log(25 - 3^n)

Solving for n, you find

n = log(25-3^n)/log(4)

And now, try the same trick, guessing the first value and using it to
calculate the next value. Everything works just fine. Your successive
calculations get you closer and closer to 2.

So what have we learned? Well, putting one of the n's on the left and
solving for it in terms of the other MIGHT be the start of a good
algorithm. If the numbers get closer and closer to something, then
it's a pretty good bet that that something is a solution to the
equation. We say the sequence "converges." But sometimes the numbers
get wilder and wilder with each re-calculation.  We say the sequence
"diverges."

Can we predict in advance which is going to happen? Are there other
possibilities, besides converging and diverging?

Lots of good things to try and to think about... Please keep in touch.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Sequences, Series

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