infinite descent

Fermat invented this method of infinite descent. The idea is: If a given natural numberMathworldPlanetmath 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 coprimeMathworldPlanetmath positive integers with opposite parity, m<n, and, say, m is even.

Let a=2mn, b=n2-m2, c=m2+n2. Then {a,b,c} is a primitive Pythagorean tripleMathworldPlanetmath, and the area A of the right triangleMathworldPlanetmath with sides a,b,c is ab/2=mn(n2-m2).

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 r2=m, s2=n, we have A=(rs)2(s4-r4).

We prove that the Diophantine equationMathworldPlanetmath x4-y4=z2 has no solution in natural numbers.

Remark 1.

Suppose that z2+y4=x4, where gcd(x,y,z)=1, x,y,zN. Then x is odd, and y,z have opposite parity.


If x was even, then x4=z2+y4(z+y2)20(mod2), so z,y20(mod2) or z,y21(mod2). But z,y20(mod2) conflicts with gcd(x,y,z)=1. And z,y21(mod2) implies y2+(z2)22(mod4) contradicting x40(mod4). Thus, x is odd, and x4=z2+(y2)2(z+y2)21(mod2) implies that z,y2 have opposite parity. ∎

Suppose x is odd and z is even. Then we have z=2pq, y2=q2-p2 and x2=q2+p2, where p,q have opposite parity and are coprime. Since z is odd, this implies (xy)2=q4-p4, 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, m<n,(2mn,m+n)=1 such that

y2 =2mn (1)
x2 =n2 +m2 (2)
z =n2 -m2. (3)

Since m2+n2=x2 is a primitive Pythagorean triple, there exist p,q, p<q, (2pq,p+q)=1 satisfiying

m =2pq (4)
n =q2 -p2 (5)
x =q2+p2. (6)

Since 2mn is a square and m,n are coprime and, say, n is odd, n is a square, and we have m=2r2, n=s2.

From the primitive Pythagorean triple m2+n2=x2 we get x=u2+v2, n=u2-v2, m=2uv. Since 2uv=2r2 uv is a square, and each of u and v is a square: u=g2, v=h2.

Substituting n,u,v in n=u2-v2 we have s2=g4-h4. But since n+(1/2)<m this implies n=s2<z=n2-m2<z2, thus we have another solution with odd s<z. This contradicts to the fact that there exists a smallest solution.

See for a discussion of infinite descent vs. inductionMathworldPlanetmath.

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