PlanetMath: infinite descent

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infinite descent

(Topic)

Fermat invented this method of infinite descent. The idea is: If a given natural number 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 be coprime positive integers with opposite parity, , and, say, is even.

Let , , . Then $\{a,b,c\}$ is a primitive Pythagorean triple, and the area of the right triangle with sides is .

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

We prove that the Diophantine equation has no solution in natural numbers.

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

Proof. If

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,

is odd, and $x^4=z^2+(y^2)^2 \equiv (z+y^2)^2 \equiv 1 \pmod{2}$ implies that

have opposite parity. $\qedsymbol$

Suppose

is odd and

is even. Then we have

and

, where

have opposite parity and are coprime. Since

is odd, this implies

, so it is sufficient to show that there is no solution for odd

Now are assumed odd. Then is even, and there exist $m,n \in {\mathbb{N}}$ , , 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

is a primitive Pythagorean triple, there exist $p,q \in {\mathbb{N}}$ ,

satisfiying

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

Since

is a square and

are coprime and, say,

is odd,

is a square, and we have

From the primitive Pythagorean triple we get , , . Since is a square, and each of and is a square: , .

Substituting in we have . But since this implies , thus we have another solution with odd . This contradicts to the fact that there exists a smallest solution.