# Monday Math: Infinite Descent

I know what you’re thinking. You’re thinking, “Gosh, it sure is neat that we can generate all Pythagorean triples from one simple formula, but what happens if we try an exponent bigger than two? That is, can you find nontrivial integer solutions to the equation

$x^n+y^n=z^n$

when n is a positive integer bigger than two? We say “nontrivial” to avoid silly situations where, for example, all three variables are equal to zero.

This, of course, is the famous Fermat equation. Our question was finally answered, by Andrew Wiles in the mid-nineties, in the negative. Sadly, his proof is way too complicated to discuss here. We can, however, discuss a special case. We will show that the equation

$x^4+y^4=z^2$

has no nontrivial integer solutions. Of course, since any perfect fourth power is also a perfect square, this equation is more general than the Fermat equation for exponent four.

Even cooler than the result itself is the method we are going to use to prove it. It is called “infinite descent” which, be honest, sounds almost unbearably cool don’t you think? The strategy is this: We are going to assume that there is a solution x, y, z to the equation. We are then going to manufacture a second solution X, Y, Z in which Z is smaller than z. If we are successful in that, then we will effectively be done. For we could then repeat the procedure to produce a smaller solution still, and keeping doing that as many times as we want. But since all of our variables represent positive integers, this descent cannot continue forever. We will have reached a contradiction. Let us begin.

Let x,y,z be a solution to our equation. We shall assume that the three numbers have no divisor in common. (The case where they do have a divisor in common must be handled separately, but we will skip that detail). Notice that we have:

$(x^2)^2+(y^2)^2=z^2,$

which implies that

$x^2, \ y^2, \ z$

is a primitive Pythagorean triple. In any such triple the legs are of different parity while the hypotenuse is odd. So let us assume that $x^2$ is even and $y^2$ is odd.

Using a small variation on the formula we derived last week, we can now find relatively prime positive integers u, v such that

$x^2=2uv, \phantom{xxx} y^2=u^2-v^2, \phantom{xxx} \textrm{and} \phantom{xxx} z=u^2+v^2.$

Look at that middle equation. We can rearrange it to get

$y^2+v^2=u^2,$

which means we have a second primitive Pythagorean triple y,v,u with y still odd and v even. So let’s repeat the process! This time we will let r, s denote our relatively prime integers, and now we have:

$v=2rs, \phantom{xxx} y=r^2-s^2, \phantom{xxx} \textrm{and} \phantom{xxx} u=r^2+s^2.$

Since u, v are relatively prime, it follows that

$gcd \left( u, \frac{v}{2} \right)=1,$

which is to say that these numbers are relatively prime as well. Next, notice that we have

$\left(\frac{x}{2} \right)^2=u \left(\frac{v}{2} \right).$

Thus, the product of two relatively prime integers produces a perfect square. That is only possible if both factors are perfect squares to begin with. So let us set:

$u=Z^2 \phantom{xxx} \textrm{and} \phantom{xxx} \frac{v}{2}=W^2.$

We can repeat this logic with the equation

$rs=\frac{v}{2}=W^2,$

to conclude that r ands are both perfect squares. This time we shall write

$r=X^2 \phantom{xxx} \textrm{and} \phantom{xxx} s=Y^2.$

Time for the big finish. Notice that we have

$X^4+Y^4=r^2+s^2=u =Z^2,$

which tells us that X, Y, Z is another solution to our equation. But we also have

$z=u^2+v^2=Z^4+v^2,$

which is clearly greater than Z.

Success! We produced a second solution to our equation with a smaller third number than we started with. We could now repeat this procedure again and again to produce the promised infinite descent. But since the positive integers have an absolute lower bound, infinite descent is impossible. The proof is complete.

1. #1 Stephen Lucas
January 31, 2011

Very pretty. Infinite descent is very cool when it works, and is the basis of my favorite proof that sqrt(n) is irrational, which can be found in Conway and Guy’s “Book of Numbers.” It’s much nicer than the standard sqrt(2) proof since it generalizes so well.

Assume sqrt(n)=p/q where p and q are positive integers. Then n=p^2/q^2, so nq/p = (p^2q)/(q^2p) = p/q. Since nq/p=p/q their fractional parts must also be equal, so a/p=b/q where 0

2. #2 Stephen Lucas
January 31, 2011

Curse the preview, that replaces less than and screws up the math. Let’s try that proof again:

Assume sqrt(n)=p/q where p and q are positive integers. Then n=p^2/q^2, so nq/p = (p^2q)/(q^2p) = p/q. Since nq/p=p/q their fractional parts must also be equal, so a/p=b/q where 0 less than a less than p and 0 less than b less than q. But multiplying both sides by p/b gives a/b=p/q, so a/b=sqrt(n). This fraction has smaller top and bottom than the original, and by infinite descent can get be applied infinitely often, which is impossible with integers. qed!

3. #3 snoeman
January 31, 2011

Thanks for posting the Monday Math series, Jason. It’s a great way to start the week.

4. #4 Jason Rosenhouse
January 31, 2011

Thanks for the encouragement! Glad you like the posts.