# The Basics of Infinite Descent

The simplicity of the formulation of the Infinite Descent method may be misleading:

Let $P$ be a property that integers may or may not possess. If an assumption that a positive integer $n_{0}$ has property $P$ leads to the existence of a smaller positive integer $n_{1}\lt n_{0}$ that also satisfies $P,\space$ then no positive integer has that property.

The fact is that the all-important property $P$ may not be pointed out explicitly in a problem, nor the role of the integer $n$ that may or may not possess property $P$ may be obvious from the statement of the problem.

As an example, let's consider the following [*Orchard*, #1]:

Find all solutions in integers of $x^{3} + 2y^{3} = 4z^{3}.$

The book gives two elegant solutions and I shall add a third - in my view the most straightforward.

First observe that $(0, 0, 0)$ is certainly a solution to the equation. (Furthermore, if one of $x$, $y$, or $z$ is $0$, so are the other two.) We shall see that no other solutions exist.

### Solution 1

First note that if $(x,y,z)$ is a solution in integers, then $x$ must be even, say $x=2w$. Substituting this, dividing by 2, and subtracting $y^3$, we see that $(-y)^{3}+2z^{3}=4w^{3}$, so $(-y,z,w)=(-y,z,x/2)$ is another solution. Now repeat the process to get another solution $(-z,x/2,-y/2)$ and then one more: $(-x/2,-y/2,-z/2)$. To remind, the latter is a solution in integers of the given equation, implying in particular that $x$, $y$, and $z$ are all even. But if $(x,y,z)$ were not all zero, we could keep replacing $(x,y,z)$ with $(-x/2,-y/2,-z/2)$ and eventually arrive at a solution containing an odd integer, a contradiction.

**Remark**: While there is a clear sense that the argument employs infinite descent, it may be hard to finger the property $P$ and the variable $n$ that may or may not possess that property. One possibility (that does not exactly fit the spirit of the proof) is to choose the least common multiple $n = LCM(|x|,|y|,|z|)$ of integers $x,y,z$, and declare $n$ to possess property $P$ if there are $x,y,z$ that satisfy the equation while having the $LCM$ equal to $n$. If that is the choice, it's not quite obvious why three steps in the solution were needed where one would suffice.

On the other hand, having three steps allows us to focus on any coordinate. From a solution with, say, the first coordinate $x$, we arrive at another solution with the first coordinate $-x/2$. Comparing the absolute values clearly explains the infinite descent.

### Solution 2

Suppose $(x,y,z)$ is a non-zero solution for which $|x|^{2}+2|y|^{2}+4|z|^{2}$ is minimized. Clearly $x$ is even, say $x=2w$. We have, as before, $(-y)^{3}+2x^{3}=4w^{3}$. Then $(-y,z,w)$ is a nonzero solution and

$|-y|^{3}+2|z|^{3}+4|w|^{3}=\frac{|x|^{3}+2|y|^{3}+4|z|^{3}}{2},$

a contradiction. Therefore $(0,0,0)$ is the only solution.**Remark**: The situation is more transparent now. $n=|x|^{2}+2|y|^{2}+4|z|^{2}$ and property $P$ signifies the existence of a solution $x,y,z$ for which $|x|^{2}+2|y|^{2}+4|z|^{2}=n$.

### Solution 3

For a given triple $(x,y,z)$ consider the norm $|(x,y,z)|=|x|+|y|+|z|$. We'll say that a positive integer $n\in\mathbb{N}$ has a property $P$ if there is a solution of the equation with the norm equal to $n$. Assume $N$ is the least such $n$ different from $0$. This in particular means there are $x,y,z$ that solve the equation and such that $|(x,y,z)|=n$. However, as we already saw, $(-y,z,x/2)$ is also a solution in integers, one for which

$|(-y,z,x/2)|=|x|/2 + |y| + |z|\lt |x|+|y|+|z|,$

provided, of course, that $x\ne 0$. But, as we already argued, if one of $x,y,z$ is $0$ so are the others. Hence, a contradiction.

### Solution 4

This solution - checking all possible cases modulo 2 - that leads to the infinite descent in one case and may be considered as a 1-step descent in the remaining ones. The solution has been communicated to me by Dr. Georg Fischer (Kenzingen, Germany), a developer of a symbolic math package in Java. The solution is a test case TF35 of the pacage.

$x,$ $y,$ $z$ may either be even or odd.

Case 1, all three even: $x=2a,$ $y=2b,$ $z=2c.$ Replacing and canceling a common factor of 8 yields the same polynomial with $a\lt x,$ $b\lt y,$ $c\lt z,$ that is the infinite descent.

Case 2-8, any other combination of polarity: $x=\{0|1\}+2a,$ $y=\{0|1\}+2b,$ $z=\{0|1\}+2c.$ These lead to polynomials with a factor of $2$ before all variables, and a constant as follows:

$[1+2a,0+2b,0+2c]:\; failure\; constant=1,\; vgcd=2$

$[0+2a,1+2b,0+2c]:\; failure\; constant=1,\; vgcd=2$

$[1+2a,1+2b,0+2c]:\; failure\; constant=3,\; vgcd=2$

$[0+2a,0+2b,1+2c]:\; failure\; constant=-1,\; vgcd=2$

$[1+2a,0+2b,1+2c]:\; failure\; constant=-3,\; vgcd=2$

$[0+2a,1+2b,1+2c]:\; failure\; constant=-1,\; vgcd=2$

$[1+2a,1+2b,1+2c]:\; failure\; constant=-1,\; vgcd=2$

These 7 combinations of polarities can never occur.

### References

- M. I. Krusemeyer, G. T. Gilbert, L. C. Larson,
*A Mathematical Orchard*, MAA, 2012

### What Is Infinite Descent?

- The Basics of Infinite Descent
- Fermat's Last Theorem for n = 4
- Pennies in Boxes
- Square root of 2 is irrational
- Fermat's Only Published Proof
- Ambassadors at a Round Table

|Activities| |Contact| |Front page| |Contents| |Arithmetic|

Copyright © 1996-2018 Alexander Bogomolny

69839990