Unsolvable Equations

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 
as your next guess, etc.

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/