**Question Corner and Discussion Area**

I want to prove, using mathematical induction, that there are no solutions to the equation , for positive values ofI assume thatx,y,z.

Those reading this page will likely recognize this question as the *n*=4
case of Fermat's Last Theorem. The general theorem (that, when
*n *> 2, there are no positive integer solutions to the equation
), was conjectured by Fermat hundreds of years ago
but remained unproved until just recently, when it was proved by
Andrew Wiles. The proof is highly complex and involves some deep areas
of abstract mathematics.

The *n*=4 case, however, is relatively easy to prove and was known by
Fermat. It turns out to be a little easier to prove the more general
result that there is no solution in positive integers to the equation
(that is, the sum of two fourth powers cannot even
be a perfect square, let alone a fourth power).

This is traditionally proven using the "method of infinite descent". The key to this method is the key lemma below (I'll explain later how this fact is proven).

**Key Lemma**. If where *x*, *y*, and *z* are
positive integers, then there would exist other positive integers
*u*, *v*, and *w* with , and *w *< *z*.

(For those unfamiliar with the terminology: a *lemma* is a small
theorem whose main use is in proving another, more important, theorem).

The reason this lemma implies that solutions cannot exist
is as follows.
Suppose solutions did exist. Among all solutions, pick one with the
smallest *z* value (any nonempty set of postive integers has a smallest
element). But now the lemma says there would be another solution with a still
smaller *z* value, a contradiction. Therefore, solutions cannot exist.

You can phrase this using the language of mathematical induction, but
it's more awkward. You would work by induction on *z*.
When *z*=1 there are clearly no solutions (*x* and *y* have
to be at least 1, so has to be at least 2).

Now, if it is known that there are no solutions with , you can
prove that there are no solutions with *z*=*N*+1 either: if (*x*,*y*,*z*)
were such a solution, then the lemma implies that there is
another solution (*u*,*v*,*w*) with *w *< *z*, so , contradicting
the fact that we know no such solution exists.

The above two paragraphs form the basis and induction steps for a proof by induction. But it's a little cleaner to use the method of infinite descent.

Now for the hard part: a proof of the lemma. I won't fill in all the
details (since you originally posted this in the *Discussion Topics*
section rather than the *Questions* section, I assume you don't want
me too), but I will outline the proof.

It depends on the theory of Pythagorean triples. A triple (*a*,*b*,*c*) of
positive integers is called a *Pythagorean triple* if
. It is called a *fundamental Pythagorean triple*
if *a*, *b*, and *c* have no common factor greater than 1.

For any fundamental Pythagorean triple, one of *a* and *b* is even and
the other is odd. Let's use the letter *a* to refer to the even one, *b* to
the odd one. Then there are positive integers *m* and *n* such that

and *m* and *n* are relatively prime (have no common factor greater than
1).

If you don't know how to prove the above statements, feel free to post another question here.

Now, suppose . Then is a Pythagorean triple.

If *x*, *y*, and *z* have a common prime factor *p *> 1,
then and we have our
desired solution (*u*,*v*,*w*) with *u *= *x*/*p*, *v *= *y*/*p*, and
. (You need also to explain why is an integer;
I'll leave that to you).

On the other hand, if *x*, *y*, and *z* do *not* have a common
prime factor, then is a fundamental Pythagorean triple.
One of *x* and *y* is even and the other odd; let's use *x* to denote the
even one. Then, by the theory of Pythagorean triples,
there are relatively prime positive integers *m* and *n*
such that

We can rewrite the second equation as .
We know that *m* and *n* have no common factor greater than 1, so
(*n*,*y*,*m*) is a fundamental Pythagorean triple. We know *y* is odd, so
*n* must be the even one. Therefore, appealing to the
theory of Pythagorean triples once again, there are relatively prime
positive integers *r* and *s* such that

The final piece of the puzzle is the fact that if the product of
two relatively prime positive integers is a perfect square, then
each individually is a perfect square. From this it follows that
*m* and *n*/2 are perfect squares, since the product is
(remember that
*x* and *n* are even, so *x*/2 and *n*/2 are integers).

It also follows that *r* and *s* are perfect squares, since
the product is which is a perfect square.

Therefore, setting , , ,
the equation becomes
. All that remains to complete the proof of the
lemma is to show that *w *< *z*, which follows because
, so , and as *w* is a
positive integer this implies *w *< *z* as well.

If you want proofs of some of the statements I've left for you to fill in, just post a message here and we'd be glad to fill them in for you.

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Last updated: April 19, 1999

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