Infinite descent

In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. One typical application is to show that a given equation has no solutions.

Assuming an example with a particular property exists, one shows that another exists that is in some sense 'smaller' as measured by a natural number. Then by mathematical induction (infinitely repeating the same step), one shows there are a yet smaller, then a yet even smaller, example, and hence there must be an infinitude of ever smaller examples. Since there are only a finite number of natural numbers smaller than the size of the initially postulated example, this is impossible—it is a contradiction, so no such initial example can exist.

This illustrative description can be restated in terms of a minimal counterexample, giving a more common type of proof formulation. We suppose a 'smallest' solution and then derive a smaller one — thereby getting a contradiction.

The method was developed by and much used for Diophantine equations by Fermat. Two typical examples are showing the non-solvability of the Diophantine equation r² + s⁴ = t⁴ and proving that any prime p such that p ≡ 1 (mod 4) can be expressed as a sum of two squares. In some cases, to a modern eye, what he was using was (in effect) the doubling mapping on an elliptic curve. More precisely, his method of infinite descent was an exploitation in particular of the possibility of halving rational points on an elliptic curve E by inversion of the doubling formulae. The context is of a hypothetical rational point on E with large co-ordinates. Doubling a point on E roughly doubles the length of the numbers required to write it (as number of digits): so that a 'halved' point is quite clearly smaller. In this way Fermat was able to show the non-existence of solutions in many cases of Diophantine equations of classical interest (for example, the problem of four perfect squares in arithmetic progression).

Number theory

In the number theory of the twentieth century, the infinite descent method was taken up again, and pushed to a point where it connected with the main thrust of algebraic number theory and the study of L-functions. The structural result of Mordell, that the rational points on an elliptic curve E form a finitely-generated abelian group, used an infinite descent argument based on E/2E in Fermat's style.

To extend this to the case of an abelian variety A, André Weil had to make more explicit the way of quantifying the size of a solution, by means of a height function - a concept that became foundational. To show that A(Q)/2A(Q) is finite, which is certainly a necessary condition for the finite generation of the group A(Q) of rational points of A, one must do calculations in what later was recognised as Galois cohomology. In this way, abstractly-defined cohomology groups in the theory become identified with descents in the tradition of Fermat. The Mordell-Weil theorem was at the start of what later became a very extensive theory.

Application examples

Irrationality of √2

Suppose that √2 were rational. Then it could be written as

$\sqrt{2} = \frac{p}{q}$

for two natural numbers, p and q. Then squaring would give

$2 = \frac{p^2}{q^2}$,

2q² = p²,

so

2 | p².

This implies that

2 | p,

which means that "2 divides p" (that is, "p is divisible by 2"). (If 2 did not divide p, then the prime factorization of p (the product of its primes) would contain no 2's. So when one squares p by squaring all its factors, there still would be no 2's in the resulting prime factorization of p². But since p² has been found to be divisible by 2, p must be divisible by 2 as well.)

We can now write p = 2r; thus

2q² = (2r)² = 4r²,

q² = 2r², so

2 | q² and

2 | q.

Therefore, for both p and q, smaller natural numbers (which would work equally well to form the rational) can be found by dividing them in half. The same must hold for those smaller numbers, ad infinitum. However, this is impossible in the set of natural numbers. Since √2 is a real number, which can be either rational or irrational, the only option left is for √2 to be irrational.

Irrationality of √k if it is not an integer

For positive integer k, suppose that √k is not an integer, but is rational. Then express it in lowest possible terms (i.e., as a fully reduced fraction) as ^m⁄_n for natural numbers m and n, and let q be the largest integer no greater than √k. Then

$\begin{align} \sqrt k&=\frac mn\\ &=\frac{m(\sqrt k-q)}{n(\sqrt k-q)}\\ &=\frac{m\sqrt k-mq}{n\sqrt k-nq}\\ &=\frac{nk-mq}{m-nq}\end{align}$

and therefore √k can be expressed in lower terms, which is a contradiction.

Non-solvability of r² + s⁴ = t⁴

The non-solvability of r² + s⁴ = t⁴ in integers is sufficient to show the non-solvability of q⁴ + s⁴ = t⁴ in integers, which is a special case of Fermat's Last Theorem, and the historical proofs of the latter proceeded by more broadly proving the former using infinite descent. A more recent proof demonstrates both of these impossibilities by proving still more broadly that a Pythagorean triangle cannot have two unspecified sides each of which is each either a square or twice a square, since there is no smallest such triangle.

Suppose there exists such a Pythagorean triangle. Then it can be scaled down to give a primitive (i.e., with no common factors) Pythagorean triangle with the same property. Primitive Pythagorean triangles' sides can be written as x = 2ab, y = a² − b², z = a² + b², with a and b relatively prime and with a+b odd and hence y and z both odd. There are three cases, depending on which two sides are postulated to each be a square or twice a square:

• y and z: Neither y nor z, being odd, can be twice a square; if they are both square, the right triangle with legs $\sqrt{yz}$ and b² and hypotenuse a² also would have integer sides including a square leg (b²) and a square hypotenuse (a²), and would have a smaller hypotenuse (a² compared to z = a² + b²).

• y and x: If y is a square and x is a square or twice a square, then each of a and b is a square or twice a square and the integer right triangle with legs b and $\sqrt{y}$ and hypotenuse a would have two sides (b and a) each of which is a square or twice a square, with a smaller hypotenuse than the original triangle (a compared to z = a² + b²).

• z and x: If z is a square and x is a square or twice a square, again each of a and b is a square or twice a square and the integer right triangle with legs a and b and hypotenuse $\sqrt{z}$ also would have two sides (a and b) each of which is a square or twice a square, and a smaller hypotenuse ($\sqrt{z}$ compared to z).

In any of these cases, one Pythagorean triangle with two sides each of which is a square or twice a square has led to a smaller one, which in turn would lead to a smaller one, etc.; since such a sequence cannot go on infinitely, the original premise that such a triangle exists must be wrong. This implies that r² + s⁴ = t⁴ cannot have a solution, since if it did then r, s², and t² would be the sides of such a Pythagorean triangle.

For other proofs of this by infinite descent, see and .

Non-solvability of a² + b² = 3(s² + t²)

Infinite descent can be used to show that there are no integer solutions to

$a^2+b^2=3 \cdot (s^2+t^2)$,

other than a = b = s = t = 0.

Suppose there is a nontrivial integer solution of the equation. Then there is a nontrivial nonnegative integer solution obtained by replacing each of a,b,s,t by its absolute value. So it suffices to show that there are no nontrivial nonnegative integer solutions.

Suppose that a₁,b₁,s₁,t₁ is a nonnegative solution. We have

$3 \mid a_1^2+b_1^2$

This is only true if both a₁ and b₁ are divisible by 3. Let

3a₂ = a₁ and 3b₂ = b₁.

Thus we have

$(3 a_2)^2 + (3 b_2)^2 = 3 \cdot (s_1^2+t_1^2)$

and

$3(a_2^2+b_2^2) = s_1^2+t_1^2$,

which yields a new nontrivial nonnegative integer solution s₁, t₁, a₂, b₂. Under a suitable notion of size of the solutions, e.g. the sum of the four integers, this new solution is smaller than the original one. This process can be repeated infinitely, producing an infinite decreasing sequence of positive solution sizes. This is a contradiction, because no such sequence exists. This shows that there are no nonzero solutions for this Diophantine equation.

See also

• Proof by contradiction
• Vieta jumping

Other reading

• Infinite descent at PlanetMath.
• Example of Fermat's last theorem at PlanetMath.