Proof by infinite descent

In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the fact that the natural numbers are well ordered. One typical application is to show that a given equation has no solutions. Assuming a solution exists, one shows that another exists, that is in some sense 'smaller'. Then one must show, usually with greater ease, that the infinite descent implied by having a whole sequence of solutions that are ever smaller, by our chosen measure, is an impossibility. This is a contradiction, so no such initial solution can exist.

This illustrative description can be restated in terms of a minimal counterexample, giving a more common type of formulation of an induction proof. We suppose a 'smallest' solution - then derive a smaller one. That again is a contradiction.

The method can be seen at work in one of the proofs of the irrationality of the square root of two. It was developed by and much used for Diophantine equations by Fermat. Two typical examples are solving the diophantine equation $x^{4}+y^{4}=z^{2}$ and proving a prime 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 2 integers, p and q. Then,

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

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

so $2|p$. Let $p=2P$, and

$\displaystyle {2q^{2}=(2P)^{2}=4P^{2}}$

$q^{2}=2P^{2},\,$

so $2|q$. Therefore, for both p and q, smaller integers 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 integers. Since √2 is a real number, which can be either rational or irrational, the only option left is for √2 to be irrational.

A Diophantine equation

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 $a_{1},b_{1},s_{1},t_{1}$ is a nonnegative solution. We have

$3\mid a_{1}^{2}+b_{1}^{2}$

This is only true if both $a_{1}$ and $b_{1}$ are divisible by 3. Let

$3a_{2}=a_{1}\,$ and $3b_{2}=b_{1}.\,$

Thus we have

$(3a_{2})^{2}+(3b_{2})^{2}=3\cdot (s_{1}^{2}+t_{1}^{2})$

and

$3(a_{2}^{2}+b_{2}^{2})=s_{1}^{2}+t_{1}^{2},\,$

which yields a new nontrivial nonnegative integer solution $s_{1},t_{1},a_{2},b_{2}$. 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