infinite descent InfiniteDescent 2004-02-02 11:16:50 2004-02-13 10:11:20 Topic % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\N}{{\mathbb N}} \newcommand{\Z}{{\mathbb Z}} \newtheorem{rmk}{Remark} \newtheorem{lem}{Lemma} \newtheorem{cor}{Corollary} \newtheorem{theo}{Theorem} 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. \begin{rmk} Suppose that $z^2+y^4=x^4$, where $\gcd(x,y,z)=1$, $x,y,z \in \N$. Then $x$ is odd, and $y,z$ have opposite parity. \end{rmk} \begin{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. \end{proof} 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 $m,n \in \N$, $m<n$,$(2mn,m+n)=1$ such that \begin{eqnarray} \label{eq1} y^2&=2mn \\ x^2&=n^2&+m^2 \\ z&=n^2&-m^2. \end{eqnarray} Since $m^2+n^2=x^2$ is a primitive Pythagorean triple, there exist $p,q \in \N$, $p<q$, $(2pq,p+q)=1$ satisfiying \begin{eqnarray} \label{eq2} m&=2pq \\ n&=q^2&-p^2 \\ x&=q^2+p^2. \end{eqnarray} 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 $2uv=2r^2$ $uv$ 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 \PMlinkexternal{here}{http://mathpages.com/home/kmath144.htm} for a discussion of infinite descent vs. induction.