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### Solving a^n + b^n = c^n

```Date: 03/10/2009 at 22:00:13
From: Tim
Subject: Fermat's last theorem

What if n < 2 ?  If n = -2, we have c^2 = (a^2*b^2)/(a^2 + b^2) and
there is a solution.

What is the general approach?

```

```
Date: 03/13/2009 at 14:26:59
From: Doctor Vogler
Subject: Re: Fermat's last theorem

Hi Tim,

Thanks for writing to Dr. Math.

When n is negative, say n = -m, then

c^(-m) = a^(-m) + b^(-m)

is equivalent (when abc is nonzero) to

c^(-m) = (a^m + b^m)/(ab)^m

c^m = (ab)^m/(a^m + b^m)

c^m(a^m + b^m) = (ab)^m

(ac)^m + (bc)^m = (ab)^m

and so by Fermat's Last Theorem, there are no integer solutions when m
is bigger than 2.  So the only cases left to check are n = -2, -1, 0,
1, and 2.

Of course, for n=2, we have the well-known Pythagorean triples:

Dr. Math FAQ: Pythagorean Triples
http://mathforum.org/dr.math/faq/faq.pythag.triples.html

When n=1, the solution is easy, since c = a + b implies that all
solutions have the form (a, b, a+b).

When n=0, there is no solution, because the equation becomes 1 = 2.

When n=-1, the solution is a little trickier, but not too bad.  You
end up with ac + bc = ab, which you can also factor as

(a-c)(b-c) = c^2.

So we could let r = a-c and let s = b-c, and then rs = c^2.  If we let
g = gcd(r, s), then (r/g)(s/g) = (c/g)^2, and r/g and s/g are
relatively prime integers whose product is a square.  So that means
that both r/g and s/g are squares of integers, which we will call r/g
= u^2 and s/g = v^2.  And then c/g = uv.  Therefore, all solutions
come from a relatively prime pair (u, v) and an integer g, giving the
solution

(gu(u+v), gv(u+v), guv).

Finally, we have the case that you asked about, where n = -2 and

c^2 = (ab)^2/(a^2 + b^2)

or

(ac)^2 + (bc)^2 = (ab)^2.

You can expect this is a little bit harder than the case of n=2.

If we use the method of

Rational Solutions to Two Variable Quadratic Equation
http://mathforum.org/library/drmath/view/65319.html

then it is pretty easy to determine that all rational solutions have
the form

a = 2t(t^2 + 1)k
b =   (t^4 - 1)k
c = 2t(t^2 - 1)k

for some rational numbers t and k.  So all integer solutions have the form

a = 2uv(u^2 + v^2)k
b =    (u^4 - v^4)k
c = 2uv(u^2 - v^2)k

where u and v are relatively prime integers and k is a rational
number.  But you can show that gcd(2uv(u^2 + v^2), u^4 - v^4, 2uv(u^2
- v^2)) is either 4 (if u and v are both odd) or 1.  Therefore, all
integer solutions have the form

a = 2uv(u^2 + v^2)k/4
b =    (u^4 - v^4)k/4
c = 2uv(u^2 - v^2)k/4

for some pair of relatively prime *odd* integers (u, v), and some
integer k, or

a = 2uv(u^2 + v^2)k
b =    (u^4 - v^4)k
c = 2uv(u^2 - v^2)k

for some pair of relatively prime integers (u, v), not both odd, and
some integer k.  You could probably also do a trick like for
Pythagorean triples where you eliminate the other case by switching a
and b.  That is, if we have

a = 2uv(u^2 + v^2)k/4
b =    (u^4 - v^4)k/4
c = 2uv(u^2 - v^2)k/4

with u and v both odd, then we can simply set U = (u+v)/2 and V =
(u-v)/2 so that

b = 2UV(U^2 + V^2)k
a =    (U^4 - V^4)k
c = 2UV(U^2 - V^2)k

We could probably get at the same result by factoring the equation as

a^2*c^2 = b^2(a - c)(a + c),

assuming some coprimality condition, and building up the solution like
in the link above on Pythagorean triples.

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory

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