infinite descent

Fermat invented this method of infinite descent. The idea is: If a given natural number $n$ with certain properties implies that there exists a smaller one with these properties, then there are infinitely many of these, which is impossible.

Here is an example:

Let $m, n$ be coprime positive integers with opposite parity, $m<n$ , and, say, $m$ is even.

Let $a=2mn$ , $b=n^{2}-m^{2}$ , $c=m^{2}+n^{2}$ . Then $\{a,b,c\}$ is a primitive Pythagorean triple, and the area $A$ of the right triangle with sides $a, b, c$ is $ab/2=mn(n^{2}-m^{2})$ .

Suppose $A$ is a square. Then, since $m, n$ are coprime and of opposite parity, $\gcd(m+n,m-n)=\gcd(m,n)=1$ . Thus, for $A$ to be a square, each of $m,n,m-n,m+n$ must be squares itself. Setting $r^{2}=m$ , $s^{2}=n$ , we have $A=(rs)^{2}(s^{4}-r^{4})$ .

We prove that the Diophantine equation $x^{4}-y^{4}=z^{2}$ has no solution in natural numbers.

Remark 1.

Suppose that $z^{2}+y^{4}=x^{4}$ , where $\gcd(x,y,z)=1$ , $x,y,z\in{\mathbb{N}}$ . Then $x$ is odd, and $y, z$ have opposite parity.

Proof.

If $x$ was even, then $x^{4}=z^{2}+y^{4}\equiv(z+y^{2})^{2}\equiv 0\pmod{2}$ , so $z,y^{2}\equiv 0\pmod{2}$ or $z,y^{2}\equiv 1\pmod{2}$ . But $z,y^{2}\equiv 0\pmod{2}$ conflicts with $\gcd(x,y,z)=1$ . And $z,y^{2}\equiv 1\pmod{2}$ implies $y^{2}+(z^{2})^{2}\equiv 2\pmod{4}$ contradicting $x^{4}\equiv 0\pmod{4}$ . Thus, $x$ is odd, and $x^{4}=z^{2}+(y^{2})^{2}\equiv(z+y^{2})^{2}\equiv 1\pmod{2}$ implies that $z,y^{2}$ have opposite parity. ∎

Suppose $x$ is odd and $z$ is even. Then we have $z=2pq$ , $y^{2}=q^{2}-p^{2}$ and $x^{2}=q^{2}+p^{2}$ , where $p, q$ have opposite parity and are coprime. Since $z$ is odd, this implies $(xy)^{2}=q^{4}-p^{4}$ , so it is sufficient to show that there is no solution for odd $z$ .

Now $x, z$ are assumed odd. Then $y$ is even, and there exist $m,n\in{\mathbb{N}}$ , $m<n$ , $(2mn,m+n)=1$ such that

$\displaystyle y^{2}$	$\displaystyle=2mn$		(1)
$\displaystyle x^{2}$	$\displaystyle=n^{2}$	$\displaystyle+m^{2}$	(2)
$\displaystyle z$	$\displaystyle=n^{2}$	$\displaystyle-m^{2}.$	(3)

Since $m^{2}+n^{2}=x^{2}$ is a primitive Pythagorean triple, there exist $p,q\in{\mathbb{N}}$ , $p<q$ , $(2pq,p+q)=1$ satisfiying

$\displaystyle m$	$\displaystyle=2pq$		(4)
$\displaystyle n$	$\displaystyle=q^{2}$	$\displaystyle-p^{2}$	(5)
$\displaystyle x$	$\displaystyle=q^{2}+p^{2}.$		(6)

Since $2mn$ is a square and $m, n$ are coprime and, say, $n$ is odd, $n$ is a square, and we have $m=2r^{2}$ , $n=s^{2}$ .

From the primitive Pythagorean triple $m^{2}+n^{2}=x^{2}$ we get $x=u^{2}+v^{2}$ , $n=u^{2}-v^{2}$ , $m=2uv$ . Since $2uv=2r^{2}$ $u v$ is a square, and each of $u$ and $v$ is a square: $u=g^{2}$ , $v=h^{2}$ .

Substituting $n, u, v$ in $n=u^{2}-v^{2}$ we have $s^{2}=g^{4}-h^{4}$ . But since $n+(1/2)<m$ this implies $n=s^{2}<z=n^{2}-m^{2}<z^{2}$ , thus we have another solution with odd $s<z$ . This contradicts to the fact that there exists a smallest solution.

See http://mathpages.com/home/kmath144.htmhere for a discussion of infinite descent vs. induction.

Title	infinite descent
Canonical name	InfiniteDescent
Date of creation	2013-03-22 14:07:56
Last modified on	2013-03-22 14:07:56
Owner	Thomas Heye (1234)
Last modified by	Thomas Heye (1234)
Numerical id	13
Author	Thomas Heye (1234)
Entry type	Topic
Classification	msc 11D25
Related topic	ExampleOfFermatsLastTheorem