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, thenthere are infinitely many of these, which is impossible.

Here is an example:

Let $m,n$ be coprime positive integers with opposite parity, , 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 besquares 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 $z^{\mathrm{2}}\mathrm{+}{y}^{\mathrm{4}}\mathrm{=}{x}^{\mathrm{4}}$, where $\gcd \mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{1}$, $x\mathrm{,}y\mathrm{,}z\mathrm{\in }\mathrm{N}$. Then $x$ is odd, and$y\mathrm{,}z$ have opposite parity.

Proof.

If $x$ was even, then $x^4=z^2+y^4\equiv {(z+y^2)}^2\equiv 0(mod2)$, so $z,y^2\equiv 0(mod2)$ or $z,y^2\equiv 1(mod2)$. But $z,y^2\equiv 0(mod2)$ conflicts with$\gcd (x,y,z)=1$. And $z,y^2\equiv 1(mod2)$ implies $y^2+{(z^2)}^2\equiv 2(mod4)$contradicting $x^4\equiv 0(mod4)$. Thus, $x$ is odd, and $x^4=z^2+{(y^2)}^2\equiv {(z+y^2)}^2\equiv 1(mod2)$ 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 $m,n\in \mathbb{N}$,,$(2mn,m+n)=1$ such that

$y^2$$=2mn$(1)

$x^2$$=n^2$$+m^2$(2)

$z$$=n^2$$-m^2.$(3)

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

$m$$=2pq$(4)

$n$$=q^2$$-p^2$(5)

$x$$=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 $2uv=2r^2$ $uv$ 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 this implies , thus we have another solution with odd . This contradicts to the fact that there exists a smallest solution.

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